KC Sinha Solution Class 12 Chapter 19 Indefinite Integrals (अनिश्चित समाकल) Exercise 19.13

 

 Exercise 19.13

Example-3

$\int \frac{dx}{3+2x-x^2}$

Sol :

Given : $I=\int \frac{dx}{3+2x-x^2}$

$=\int \frac{dx}{3-(x^2-2x)}$

$=\int \frac{dx}{3-(x^2-2x+1-1)}$

$=\int \frac{dx}{3-(x^2-2x+1)+1}$

$=\int \frac{dx}{4-(x^2-2x+1)}$

$=\int \frac{dx}{(2)^2-(x-1)^2}$

$=\frac{1}{2\times 2}\log \left|\frac{2+x-1}{2-x+1}\right|+C$

$\left[\therefore \int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|\right]$

$=\frac{1}{4}\log \left|\frac{x+1}{3-x}\right|+C$

Question 1

(i) $\int \frac{dx}{x^2-9}$

Sol :

$I=\int \frac{dx}{x^2-9}$

Let $\frac{1}{(x-3)(x+3)}=\frac{A}{x-3}+\frac{B}{x+3}$

$\frac{1}{(x-3)(x+3)}=\frac{A(x+3)+B(x-3)}{(x-3)(x+3)}$

1=A(x+3)+B(x-3)

Putting x=3 , we get 1=A(3+3)+B(3-3)=6A

$\therefore A=\frac{1}{6}$

Putting x=-3 we get 1=A(-3+3)+B(-3-3)

1=-6B

$B=-\frac{1}{6}$

Now , $I=\int \frac{1}{(x-3)(x+3)}dx$

$=\int \left(\frac{A}{x-3}+\frac{B}{(x+3)}\right)$

$=A\int \frac{1}{x-3}dx+B\int \frac{1}{x+3}dx$

$=\frac{1}{6}\log |x-3|-\frac{1}{6}\log |x+3|+C$

$\left[\therefore \log~m-\log~n=\log\frac{m}{n}\right]$

$=\frac{1}{6}\log \left|\frac{x-3}{x+3}\right|+C$

(ii) $\int \frac{x}{(x++1)(x+2)}dx$

Sol :

Given: $I=\int \frac{x}{(x+1)(x+2)}dx$

Let $\frac{x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$

$=\frac{x}{(x+1)(x+2)}=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}$

x=A(x+2)+B(x+1)

Putting x=-1 we get -1=A(-1+2)+B(-1+1)

-1=A

Putting x=-2 We get -2=A(-2+2)+B(-2+1)

-2=-B

B=2

Now , $I=\int \frac{x}{(x+1)(x+2)}dx=\int \left(\frac{A}{x+1}+\frac{B}{x+2}\right)$

$=A\int \frac{1}{x+1}dx+B\int \frac{1}{x+2}dx$

I=-1 log|x+1|+2log|x+2|+C

I=2log|x+2|-log|x+1|+C

(iii) $\int \frac{dx}{(x+1)(x+2)}$

Sol :

Given : $I=\int \frac{dx}{(x+1)(x+2)}$

Let $\frac{1}{(x+1)(x+2)}=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}$

1=A(x+2)+B(x+1)

Putting x=-1 We get 1=A(-1+2)+B(-1+1)

∴A=1

Putting x=-2, We get 1=A(-2+2)+B(-2+1)

-B=1

B=-1

Now $I=\int \frac{1}{(x+1)(x+2)}dx$

$=\int \left(\frac{A}{x+1}+\frac{B}{x+2}\right)$

$=A\int \frac{1}{x+1}dx+B\int \frac{1}{x+2}dx$

I=1×log|x+1|-1×log|x+2|+C

I=log|x+1|-log|x+2|+C

$\left(\therefore \log~m-\log~n=\frac{m}{n}\right)$

$I=\log \left|\frac{x+1}{x+2}\right|$

(iv) $\int \frac{2x+1}{(x+2)(x-3}dx$

Sol :

Given : $I=\int \frac{2x+1}{(x+2)(x-3}dx$

Let $\frac{2x+1}{(x+2)(x-3)}=\frac{A}{(x+2)}+\frac{B}{(x-3)}$

$\frac{2x+1}{(x+2)(x-3)}=\frac{A(x-3)+B(x+2)}{(x+2)(x-3)}$

2x+1=A(x-3)+B(x+2)

Putting x=-2 We get 2(-2)+1=A(-2-3)+B(-2+2)

-3=-5A

$A=\frac{3}{5}$

Putting x=3 We get 2×3+1=A(3-3)+B(3+2)

7=5B

$B=\frac{7}{5}$

Now $I=\int \frac{2x+1}{(x+2)(x-3)}dx$

$=\int \left(\frac{A}{x+2}+\frac{B}{x-3}\right)dx$

$=A\int \frac{1}{x+2}dx+B\int \frac{1}{x-3}dx$

$I=\frac{3}{5}\log |x+2|+\frac{7}{5}\log |x-3|+C$

Question 2

(i) $\int \frac{2x}{x^2+3x+2}dx$

Sol :

$I=\int \frac{2x}{x^2+3x+2}dx$

$=\int \frac{2x}{(x+2)(x+1)}dx$

Let $\frac{2x}{(x+2)(x+1)}=\frac{A}{(x+2)}+\frac{B}{(x+1)}$

$=\frac{2x}{(x+2)(x+1)}=\frac{A(x+1)+B(x+2)}{(x+2)(x+1)}$

2x=A(x+1)+B(x+2)

Putting x=-2 We get 2(-2)=A(-2+1)+B(-2+2)

-4=-A

A=4

Putting x=-1 We get 2(-1)=A(-1+1)+B(-1+2)

-2=B

B=-2

Now , $I=\int \frac{2x}{(x+2)(x+1)}dx$

$=\int \left(\frac{A}{(x+2)+\frac{B}{(x+1)}}\right)$

$=A\int \frac{1}{x+2}dx+B\int \frac{1}{x+1}dx$

=4log|x+2|-2log|x+1|+C

(ii) $\int \frac{5x}{(x+1)(x^2-4)}dx$

Sol :

Given :

$I=\int \frac{5x}{(x+1)(x^2-4)}dx$

$=\int \frac{5xdx}{(x+1)(x-2)(x+2)}$

(a²-b²)=(a-b)(a+b)

Let $\frac{5x}{(x+1)(x-2)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x-2)}+\frac{C}{(x+2)}$

$\frac{5x}{(x+1)(x-2)(x+2)}=\frac{A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)}{(x+1)(x-2)(x+2)}$

5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)

Putting x=-1 We get 5(-1)=A(-1-2)(-1+2)+0+0

-5=-3A

$A=\frac{5}{3}$

Putting x=2 We get 5×2=B(2+2)(2+1)+0+0

10=12B

$B=\frac{5}{6}$

Putting x=-2 We get 5(-2)=0+0+C(-2+1)(-2-2)

-10=4C

$C=\frac{-10}{4}=\frac{-5}{2}$

Now $I=\int \frac{5x}{(x+1)(x-2)(x+2)}dx$

$=\int \left(\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2}\right)dx$

$=A\int \frac{1}{x+1}dx+B\int \frac{1}{x-2}dx+C\int \frac{1}{x+2}dx$

$=\frac{5}{3}\log |x+1|+\frac{5}{6}\log|x-2|-\frac{5}{2}\log|x+2|+C$

$=\frac{5}{3}\log|x+1|-\frac{5}{2}\log|x+2|+\frac{5}{6}\log|x-2|+C$

Question 3

(i) $\int \frac{x}{(x-1)(x-2)(x-3)}dx$

Sol :

Given $I=\int \frac{x}{(x-1)(x-2)(x-3)}dx$

Let $\frac{x}{(x-1)(x-1)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$

$\frac{x}{(x-1)(x-1)(x-3)}=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}$

x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)

Putting x=1 We get

1=A(1-2)(1-3)+0+0

1=2A

$A=\frac{1}{2}$

Putting x=2 We get

2=0+B(2-1)(2-3)+0

2=-B

B=-2

Putting x=3 We get 

3=0+0+C(3-1)(3-2)

3=2C

$C=\frac{3}{2}$

Now $I=\int \frac{x}{(x-1)(x-2)(x-3)}dx$

$=\int \left(\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\right)dx$

$=A\int \frac{1}{x-1}dx+B\int \frac{1}{x-2}dx+C\int \frac{1}{x-3}dx$

$=\frac{1}{2}\log|x-1|-2\log|x-2|+\frac{3}{2}\log|x-3|+C$

(ii) $\int \frac{2x-3}{(x^2-1)(2x+3)}dx$

Sol :

Given : $I=\int \frac{2x-3}{(x^2-1)(2x+3)}dx$

$=\int \frac{2x-3}{(x-1)(x+1)(2x+3)}dx$

Let $\frac{2}{(x-1)(x+1)(2x+3)}=\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{C}{(2x+3)}$

$=\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)}{(x-1)(x+1)(2x+3)}$

2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)

Putting x=1 We get 

2×-3=A(1+1)(2×1+3)+0+0

-1=10A

$A=\frac{-1}{10}$

Putting x=-1 We get 2(-1)-3=0+B(-1-1)(-2+3)+0

-5=-2B

$B=\frac{5}{2}$

Putting $x=\frac{-3}{2}$ We get $2\times \left(\frac{-3}{2}\right)-3=0+0+C\left(\frac{-3}{2}-1\right)\left(\frac{-3}{2}+1\right)$

$-6=C\left(\frac{-5}{2}\right)\times \left(\frac{-1}{2}\right)$

∴$C=\frac{-6\times 4}{5}=\frac{-24}{5}$

Now $I=\int \frac{2x-3}{(x-1)(x+1)(2x+3)}dx$

$=\int \left(\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3}\right)$

$=A\int \frac{1}{x-1}dx+B\int \frac{1}{x+1}dx+C\int \frac{1}{2x+3}dx$

$=-\frac{1}{10}\log|x-1|+\frac{5}{2}\log|x+1|-\frac{12}{5}\log|2x+3|+C$

Question 4

(i) $\int \frac{3x-1}{(x-1)(x-2)(x-3)}dx$

Sol :

Given : 

$I=\int \frac{3x-1}{(x-1)(x-2)(x-3)}dx$

Let $\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$

$\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}$

3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)

Putting x=1 We get 

3×1-1=A(1-2)(1-3)+0+0

2=2A

A=1

Putting x=2 We get 

3×2-1=0+B(2-1)(2-3)+0

5=-B

B=-5

Putting x=3 We get 

3×3=0+0+C(3-1)(3-2)

8=2C

C=4

Now $I=\int \frac{3x-1}{(x-1)(x-2)(x-3)}dx$

$=\int \left(\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\right)dx$

$=A\int \frac{1}{x-1}dx+B\int \frac{1}{x-2}dx+C\int \frac{1}{x-3}dx$

I=1×log|x-1|-5log|x-2|+4log|x-3|+C

I=log|x-1|-5log|x-2|+4log|x-3|+C

(ii) $\int \frac{x}{(x-1)(x-2)}dx$

Sol :

Given : $I=\int \frac{x}{(x-1)(x-2)}dx$

Let $\frac{x}{(x-1)(x-2)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}$

$\frac{x}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$

x=A(x-2)+B(x-1)

Putting x=1 We get 

1=A(1-2)+B(1-1)

1=-A

A=-1

Putting x=2 We get 

2=A(2-2)+B(2-1)

1=B

B=2

Now $I=\int \frac{x}{(x-1)(x-2)}dx$

$=\int \left(\frac{A}{x-1}+\frac{B}{x-2}\right)$

$=A\int \frac{1}{x-1}dx+B\int \frac{1}{x-2}dx$

=-1×log|x-1|+2log|x-2|+C

=2log|x-2|-log|x-1|+C

Question 5

(i) $\int \frac{\cos x}{(1+\sin x)(2+\sin x)}dx$

Sol :

Given : 

$I=\int \frac{\cos x}{(1+\sin x)(2+\sin x)}dx$

Let z=sinx then dz=cosxdx

Now $I=\int \frac{\cos xdx}{(1+\sin x)(2+\sin x)}$

$\int \frac{dz}{(1+z)(2+z)}$

Let $\frac{1}{(1+z)(2+z)}=\frac{A}{(1+z)}+\frac{B}{(2+z)}$

$\frac{1}{(1+z)(2+z)}=\frac{A(2+z)+B(1+z)}{(1+z)(2+z)}$

$\frac{1}{(1+z)(2+z)}=\frac{A(2+z)+B(1+z)}{(1+z)(2+z)}$

1=A(2+z)+B(1+z)

Putting z=-1 We get 

1=A(2-1)+B(1-1)

1=A

A=1

Putting z=-2 We get 

1=A(2-2)+B(1-2)

1=-B

B=-1

Now , $I=\int \frac{1}{(1+z)(2+z)}dz$

$=\int \left(\frac{A}{1+z}+\frac{B}{2+z}\right)dz$

$=A\int \frac{1}{1+z}dz+B\int \frac{1}{2+z}dz$

I=1×log|1+z|-1×log|2+z|+C

I=log|1+z|-log|2+z|+C

$I=\log\left|\frac{1+z}{2+z}\right|+C$ $\therefore \log~m-\log ~n=\log \frac{m}{n}$

$I=\log\left|\frac{1+\sin x}{2+\sin x}\right|$

(ii) $\int \frac{\cos x}{(1-\sin x)(2-\sin x)}dx$

Sol :

Given :

$I=\int \frac{\cos x}{(1-\sin x)(2-\sin x)}dx$

Let z=sinx then dz=cosxdx

Now $I=\int \frac{\cos x}{(1-\sin x)(2-\sin x)}dx$

$=\int \frac{dz}{(1-z)(2-z)}$

Let $\frac{1}{(1-z)(2-z)}=\frac{A}{(1-z)}+\frac{B}{(2-z)}$

$\frac{1}{(1-z)(2-z)}=\frac{A(2-z)+B(1-z)}{(1-z)(2-z)}$

1=A(2-z)+B(1-z)

Resolving into partial fraction We get

A=1 and B=-1

Now $I=\int \frac{1}{(1-z)(2-z)}dz$

$=\int \left(\frac{A}{1-z}+\frac{B}{2-z}\right)dz$

$=A\int \frac{1}{1-z}dx+B\int \frac{1}{2-z}dz$

$=\frac{1\times \log|1-z|}{(-1)}-\frac{1\times \log|2-z|}{(-1)}+C$

I=-log|1-z|+log|2-z|+C

I=log|2-sinx|-log|-sinx|+C

$I=\log\left|\frac{2-\sin x}{1-\sin x}\right|+C$

(iii) $\int \frac{x+1}{x(1+xe^{x})}dx$

Sol :

Given : $I=\int \frac{x+1}{x(1+xe^{x})}dx$

Let z=xe^x also $x=\frac{z}{e^x}$ and 

then dz=(xe^x+e^x1)dx

dz=e^x(x+1)dx

(x+1)dx=$\frac{dz}{e^x}$

Now $I=\int \frac{x+1}{z(1+xr^x)}$

$=\int \dfrac{dz}{e^x\frac{z}{e^z}(1+z)}$

$I=\int \frac{dz}{z(1+z)}$

Let $\frac{1}{z(1+z)}=\frac{A}{z}+\frac{B}{1+z}$

$\frac{1}{z(1+z)}=\frac{A(1+z)+Bz}{z(1+z)}$

1=A(1+z)+Bz

Putting z=0 and z=-1 We get

A=1 and B=-1

Now $I=\int \frac{1}{z(1+z)}dz$

$=\int \left(\frac{A}{z}+\frac{B}{1+z}\right)dz$

I=Alogz+Blog|1+z|+C

I=logz-log|1+z|+C

$I=\log\left|\frac{z}{1+z}\right|+C$

$=\log \left|\frac{xe^x}{1+x.e^x}\right|+C$

(iv) $\int \frac{\sin x}{(1-\cos x)(2-\cos x)}dx$

Sol :

Given : $I=\int \frac{\sin x}{(1-\cos x)(2-\cos x)}dx$

Let z=cosx then 

dz=-sinxdx

-dz=sinxdx

Now $I=\int \frac{\sin x dx}{(1-\cos x)(2-\cos x)}$

$=\int \frac{-dz}{(1-z)(2-z)}$

Let $\frac{-1}{(1-z)(2-z)}=\frac{A}{1-z}+\frac{B}{2-z}$

$\frac{-1}{(1-z)(2-z)}=\frac{A(2-z)+B(1-z)}{(1-z)(2-z)}$

-1=A(2-z)+B(1-z)...(i)

Putting z=1 and z=2 in eq-(i)  We get

A=-1 and B=1

Now $=\int \frac{-dz}{(1-z)(2-z)}$

$=\int \left(\frac{A}{1-z}+\frac{B}{2-z}\right)dz$

I=-Alog|1-z|-Blog|2-z|+C

I=-(-1)log|1-z|-1log|2-z|+C

I=log|1-cosx|-log|2-cosx|+C

$I=\log\left|\frac{1-\cos x}{2-\cos x}\right|+C$

Question 6

$\int \frac{\sec^2}{(2+\tan x)(3+\tan x)}dx$

Sol :

Given : $I=\int \frac{\sec^2}{(2+\tan x)(3+\tan x)}dx$

Let z=tan x then dz=sec²xdx

Now $I=\int \frac{sec^2 x dx}{(2+\tan x)(3+\tan x)}$

$=\int \frac{dz}{(2+z)(3+z)}$

Let $\frac{1}{(2+z)(3+z)}=\frac{A}{2+z}+\frac{B}{3+z}$

$\frac{1}{(2+z)(3+z)}=\frac{A(3+z)+B(2+z)}{(2+z)(3+z)}$

1=A(3+z)+B(2+z)..(i)

Putting z=-2 and z=-3 in eq-(i) We get

A=1 and B=-1

Now $I=\int \frac{1}{(2+z)(3+z)}dz$

$=\int \left(\frac{A}{2+z}+\frac{B}{3+z}\right)$

$=A\int \frac{1}{2+z}dz+B\int \frac{1}{3+z}dz$

I=1×log|2+z|-1×log|3+z|+C

I=log|2+tanx|-log|3+tanx|+C

$\therefore \left(\log m-\log n=\log \frac{m}{n}\right)$

$I=\log\left|\frac{2+\tan x}{3+\tan x}\right|+C$

Question 7

(i) $\int \frac{dx}{x(x^5+1)}$

Sol :

Let z=x⁵ then dz=5x⁴dx

 ∴$x=z^{\frac{1}{5}}$ ∴ $dx=\frac{dz}{5x^4}=\frac{dz}{5z^{\frac{4}{5}}}$

$x^4=z^{\frac{4}{5}}$ 

Now $I=\int \frac{dx}{x(x^5+1)}$

$=\int \frac{dz}{5z^{4/5}.z^{\frac{1}{5}}(z+1)}$

Let $\frac{1}{z(z+1)}=\frac{A}{z}+\frac{B}{(z+1)}$

$\frac{1}{z(z+1)}=\frac{A(z+1)+Bz}{z(z+1)}$

1=A(z+1)+Bz...eq-(i)

Putting z=0 and z=1 in eq-(i) We get

A=1 and B=-1

Now $I=\frac{1}{5}\int \frac{dz}{z(z+1)}$

$=\frac{1}{5}\int \left(\frac{A}{z}+\frac{B}{z+1}\right)dz$

$I=\frac{1}{5}A\int \frac{1}{z}dz+\frac{1}{5}\int B\int \frac{1}{z+1}dz$

$I=\frac{1}{5}\log z-\frac{1}{5}\log|z+1|+C$

$I=\frac{1}{5}\log z-\frac{1}{5}\log|z+1|+C$

$I=\frac{1}{5}\left[\log z-\log|z+1|\right]+C$

$I=\frac{1}{5}\log\left|\frac{z}{z+1}\right|+C$

$I=\frac{1}{5}\log\left|\frac{x^5}{x^5+1}\right|+C$

(ii) $\int \frac{dx}{x(x^6+1)}$

Sol :

Let x⁶=z then 

6x⁶dx=dz

$dx=\frac{dz}{6x^5}dx$

Now $I=\int \frac{dx}{x(x^6+1)}$

$=\int \frac{dz}{6x^4.x(z+1)}$

$=\int \frac{dz}{6x^6(z+1)}$

$=\frac{1}{6}\int \frac{dz}{z(z+1)}$

Let $\frac{1}{z(z+1)}=\frac{A}{z}+\frac{B}{z+1}$

$\frac{1}{z(z+1)}=\frac{A(z+1)+Bz}{z(z+1)}$

1=A(z+1)+Bz

Putting z=0 and z=-1 in eq-(i) We get

A=1 and B=-1

Now $I=\frac{1}{6}\int \frac{dz}{z(z+1)}$

$=\frac{1}{6}\int \left(\frac{A}{z}+\frac{B}{z+1}\right)dx$

$=\frac{1}{6}\left[A\int \frac{dz}{z}+B\int \frac{dz}{z+1}\right]$

$=\frac{1}{6}\left[\log z-\log|z+1|\right]$

$=\frac{1}{6}\log \left[\frac{z}{z+1}\right]+C$

$=\frac{1}{6}\log \left|\frac{x^6}{x^6+1}\right|+C$

(iii) $\int \frac{dx}{x(x^4-1)}$

Sol :

Let x⁴=z then 

4x³dx=dz

$dx=\frac{dz}{4x^3}$

Now $I=\int \frac{dx}{x(x^4-1)}$

$=\int \frac{dz}{4x^3.x(z-1)}$

$=\int \frac{dz}{4x^4(z-1)}$

$=\frac{1}{4}\int \frac{dz}{z(z-1)}$

Let $\frac{1}{z(z-1)}=\frac{A}{z}+\frac{B}{z-1}$

$\frac{1}{z(z-1)}=\frac{A(z-1)+Bz}{z(z-1)}$

1=A(z-1)+Bz..(i)

Putting z=0 and z=1 in eq-(i) We get

A=-1 and B=1

Now $I=\frac{1}{4}\int \frac{dz}{z(z-1)}$

$=\frac{1}{4}\int \left(\frac{A}{z}+\frac{B}{z-1}\right)$

$=\frac{1}{4}\left[A\log z+B \log|z-1|\right]+C$

$=\frac{1}{4}\left[-\log z+\log|z-1|\right]+C$

$=\frac{1}{4}[log|z-1|-log z]+C$

$=\frac{1}{4}[log|x^4-1|-\log x^4]+C$

$=\frac{1}{4}\log \left|\frac{x^4-1}{x^4}\right|+C$

(iv) $\int \frac{dx}{x(x^4+1)}$

Sol :

Let z=x⁴ then 

dz=4x³dx

$dx=\frac{dz}{4x^3}$

Now $I=\int \frac{dx}{x(x^4+1)}$

$=\int \frac{dz}{4x^3.x(z+1)}$

$=\frac{1}{4}\int \frac{dz}{x^4(z+1)}$

$=\frac{1}{4}\int \frac{dz}{z(z+1)}$

Let $\frac{1}{z(z+1)}=\frac{A}{z}+\frac{B}{z+1}$

$\frac{1}{z(z+1)}=\frac{A(z+1)+Bz}{z(z+1)}$

1=A(z+1)+Bz...(i)

Putting z=0 and z=-1 in eq-(i) We get

A=1 and B=-1

Now $I=\frac{1}{4}\int \frac{dz}{z(z+1)}$

$=\frac{1}{4}\int \left(\frac{A}{z}+\frac{B}{z+1}\right)dz$

$=\frac{1}{4}\left[A\log z+B\log|z+1| \right]+C$

$=\frac{1}{4}[log z-log|z+1|]+C$

$=\frac{1}{4}\log \left|\frac{z}{z+1}\right|+C$

$=\frac{1}{4}\log \left|\frac{x^4}{x^4+1}\right|+C$

Question 8

(i) $\int \frac{dx}{e^x-1}$

Sol :

Given : $I=\int \frac{dx}{e^x-1}$

$=\int \frac{e^x}{e^x(e^x-1)}dx$

Let e^x=z then e^xdx=dz

Now $I=\int \frac{e^x}{e^x(e^x-1)}dx$

$=\int \frac{dz}{z(z-1)}$

Let $\frac{1}{z(z-1)}=\frac{A}{z}+\frac{B}{z-1}$

$=\frac{A(z-1)+Bz}{z(z-1)}$

1=A(z-1)+Bz..(i)

Putting z=0 and z=1 in eq-(i) , We get

A=-1 and B=1

Now $I=\int \frac{dz}{z(z-1)}$

$=\int \left(\frac{A}{z}+\frac{B}{z-1}\right)dz$

$=A\int \frac{1}{z}dz+B\int \frac{1}{z-1}dz$

=-1×logz+1×log|z-1|+C

=-log e^x+log|e^x-1|+C

(∴ loge^x=x)

=-x+log|e^x-1|+C

(ii) $\int \frac{e^x}{(1+e^x)(2+e^x)}dx$

Sol :

Given : $I=\int \frac{e^x}{(1+e^x)(2+e^x)}dx$

Let z=e^x then e^xdx=dz

Now $I=\int \frac{e^x}{(1+e^x)(2+e^x)}dx$

$=\int \frac{dz}{(1+z)(2+z)}$

Let $\frac{1}{(1+z)(2+z)}=\frac{A}{(1+z)}+\frac{B}{(2+z)}$

$\frac{1}{(1+z)(2+z)}=\frac{A(2+z)+B(1+z)}{(1+z)(2+z)}$

1=A(2+z)+B(1+z)..(i)

Putting z=-1 and z=-2 in eq-(i) We get

A=1 and B=-1

Now $I=\int \frac{dz}{(1+z)(2+z)}$

$=\int \left(\frac{A}{1+z}+\frac{B}{2+z}\right)$

$=A\int \frac{1}{1+z}dz+B\int \frac{1}{2+z}dz$

I=1×log|1+z|-1×log|2+z|+C

I=log|1+e^x|-log|2+e^x|+C

$I=\log\left|\frac{1+e^x}{2+e^x}\right|+C$

Question 9

$\int \frac{x}{(x^2+a^2)(x^2+b^2)}dx$

Sol :

Let x²=z then

2xdx=dz

$xdx=\frac{dz}{2}$

Now $I=\int \frac{xdx}{(x^2+a^2)(x^2+b^2)}$

$=\int \frac{dz}{2(z+a^2)(z+b^2)}$

$=\frac{1}{2}\int \frac{dz}{(z+a^2)(z+b^2)}$

Let $\frac{1}{(z+a^2)(z+b^2)}=\frac{A}{(z+a^2)}+\frac{B}{(z+b^2)}$

$\frac{1}{(z+a^2)(z+b^2)}=\frac{A(z+b^2)+B(z+a^2)}{(z+a^2)(z+b^2)}$

1=A(z+b²)+B(z+a²)

Putting z=-a² We get

1=A(-a²+b²)+0

1=-A(a²-b²)

∴$A=-\frac{1}{(a^2-b^2)}$

Putting z=-b² We get

1=0+B(-b²+a²)

1=B(a²-b²)

∴$B=\frac{1}{(a^2-b^2)}$

Now $I=\frac{1}{2}\int \frac{dz}{(z+a^2)(z+b^2)}$

$=\frac{1}{2}\int \left(\frac{A}{z+a^2}+\frac{B}{z+b^2}\right)dz$

$=\frac{1}{2}A\int \frac{1}{z+a^2}dz+\frac{1}{2}B\int \frac{1}{z+b^2}dz$

$=\frac{1}{2}\times \frac{-1}{(a^2-b^2)}\log|z+a^2|+\frac{1}{2}\times \frac{1}{(a^2-b^2)}\log|z+b^2|+C$

$=\frac{1}{2(a^2-b^2)}\log|z+b^2|-\frac{1}{2(a^2-b^2)}\log|z+a^2|+C$

$=\frac{1}{2(a^2-b^2)}\left[\log|x^2+b^2|-\log|x^2+a^2|\right]+C$

$I=\frac{1}{2(a^2-b^2)}\log\left|\frac{x^2+b^2}{x^2+a^2}\right|+C$

Question 10

(i) $\int \frac{dx}{\sin x(3+2\cos x)}$

Sol :

Given : $I=\int \frac{dx}{\sin x(3+2\cos x)}$

$=\int \frac{\sin xdx}{\sin x.\sin x(3+2\cos x) }$

$=\int \frac{\sin xdx}{\sin ^2 x(3+2\cos x)}$

$=\int \frac{\sin xdx}{(1-\cos^2x)(3+2\cos x)}$

$=\int \frac{\sin xdx}{(1-\cos x)(1+\cos x)(3+2\cos x)}$

Let cosx=z then 

-sinxdx=dz 

sinxdx=-dz

Now $I=\int \frac{-dz}{(1-z)(1+z)(3+2z)}$

Let $\frac{1}{(1-z)(1+z)(3+2z)}=\frac{A}{(1-z)}+\frac{B}{(1+z)}+\frac{C}{(3+2z)}$

$\frac{1}{(1-z)(1+z)(3+2z)}=\frac{A(1+z)(3+2z)+B(1-z)(3+2z)+C(1-z)(1+z)}{(1-z)(1+z)(3+2z)}$

1=A(1+z)(3+2z)+B(1-z)(3+2z)+C(1-z)(1+z)

Putting z=1 We get 

1=A(1+1)(3+2×1)+0+0

1=A×2×5=10A

∴$A=\frac{1}{10}$

Putting z=1 We get 

1=0+B(1-(-1)(3-2)+0

1=2B

$B=\frac{1}{2}$

Putting $z=\frac{-3}{2}$ We get 

1=0+0+$C\left(1+\frac{3}{2}\right)\left(1-\frac{3}{2}\right)$

$1=C\times \frac{5}{2}\times \frac{-1}{2}$

$C=-\frac{4}{5}$

Now $I=-\int \frac{dz}{(1-z)(1+z)(3+2z)}$

$=-\int \left(\frac{A}{1-z}+\frac{B}{1+z}+\frac{C}{3+2z}\right)$

$=-A\int \frac{1}{1-z}dz-B\int \frac{1}{1+z}dz-C\int \frac{1}{3+2z}dz$

$=-\frac{1}{10}\frac{\log|1-z|}{-1}-\frac{1}{2}\log|1+z|-\left(-\frac{4}{5}\right)\frac{\log |3+2z|}{2}$

$=\frac{}1{10}\log|1-\cos x|-\frac{1}{2}\log |1+\cos x|+\frac{2}{5}\log|3+2\cos x|+C$

(ii) $\int \frac{dx}{1+3e^x+2e^{2x}}$

Sol :

Given : $I=\int \frac{dx}{1+3e^x+2e^{2x}}$

Let e^x=z then e^xdx=dz

∴dx=$\frac{dz}{e^x}=\frac{dz}{z}$

Now $I=\int \frac{dz}{z(1+3z+2z^2)}$

$=\int \frac{dz}{z(2z^2+2z+z+1)}$

$=\int \frac{dz}{z[2z(z+1)+1(z+1)]}$

$=\int \frac{dx}{z(z+1)(2z+1)}$

Let $\frac{1}{z(z+1)(2z+1)}=\frac{A}{z}+\frac{B}{z+1}+\frac{C}{2z+1}$

$\frac{1}{z(z+1)(2z+1)}=\frac{A(z+1)(2z+1)+B(z)(2z+1)+C(z)(z+1)}{z(z+1)(2z+1)}$

1=A(z+1)(2z+1)+B(z)(2z+1)+C(z)(z+1)

Putting z=0 We get

1=A(0+1)(2×0+1)+0+0

1=A

A=1

Putting z=-1 We get

1=0+B(-1)(2×(-1)+1)+0

1=B(-1)(-1)

B=1

Putting $z=-\frac{1}{2}$ We get 

1=0+0+$C\left(\frac{-1}{2}\right)\left(1-\frac{1}{2}\right)=\frac{-C}{4}$

C=-4

Now $I=\int \frac{dz}{z(z+1)(2z+1)}$

$=\int \left(\frac{A}{z}+\frac{B}{z+1}+\frac{C}{2z+1}\right)dz$

$=a\int \frac{1}{z}dz+B\int \frac{1}{z+1}dz+C\int \frac{1}{2z+1}dz$

I=1×logz+1×log|z+1|-$\frac{4\times log|2z+1|}{2}+C$

I=log e^x+log|1+e^x|-2log|1+2e^x|+C

I=x+log|1+e^x|-2log|1+2e^x|+C

Question 11

$\int \frac{3x-1}{(x-2)^2}dx$

Sol :

Given : $I=\int \frac{3x-1}{(x-2)^2}dx$

Let $\frac{3x-1}{(x-2)^2}=\frac{A}{(x-2)}+\frac{B}{(x-2)^2}$

$\frac{3x-1}{(x-2)^2}=\frac{A(x-2)+B}{(x-2)^2}$

3x-1=A(x-2)+B

Putting x=2 We get

3×2-1=A(2-2)+B

B=5

Putting x=1 We get

3×1-1=A(1-2)+5

2=-A+5

A=5-2=3

Now $I=\int \frac{3x-1}{(x-2)^2}dx$

$=\int \left(\frac{A}{x-2}+\frac{B}{(x-2)^2}\right)dx$

$=A\int \frac{1}{x-2}dx+B\int \frac{1}{(x-2)^2}dx$

=3log|x-2|+5$\int \frac{1}{(x-2)^2}dx$

Let x-2=z then dx=dz

Now I=3log|x-2|+$5\int \frac{1}{z^2}dz$

I=3log|x-2|-$\frac{5}{z}+C$

I=3log|x-2|-$\frac{5}{x-2}+C$

Question 12

(i) $\int \frac{x}{(x-1)^2(x+2)}dx$

Sol :

Given : $I=\int \frac{x}{(x-1)^2(x+2)}dx$

Let $\frac{x}{(x-1)^2(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+2)}$

$\frac{x}{(x-1)^2(x+2)}=\frac{A(x-1)(x+2)+B(x+2)+C(x-1)^2}{(x-1)^2(x+2)}$

x=A(x-1)(x+2)+B(x+2)+C(x-1)²

Putting x=-2 We get

-2=0+0+C(-2-1)²

-2=C9

$C=-\frac{2}{3}$

Putting x=1 We get

1=0+B(1+2)+0

1=3B

$B=\frac{1}{3}$

Putting x=2 We get 2=A(2-1)(2+2)+$\frac{1}{3}(2+2)+C\times 1$

2=4A$+\frac{4}{3}-\frac{2}{3}=4A+\frac{12-2}{9}$

$2=4A+\frac{10}{9}$

$4A=2-\frac{10}{9}$

$4A=\frac{18-10}{9}$

$A=\frac{8}{9}\times \frac{1}{4}=\frac{2}{9}$

Now $I=\int \frac{x}{(x-1)^2(x+2)}dx$

$=\int \left(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}\right)$

$=A\int \frac{1}{x-1}dx+B\int \frac{1}{(x-1)^2}dx+C\int \frac{1}{x+2}dx$

$=\frac{2}{9}\log|x-1|+\frac{1}{3}\times \left(\frac{-1}{x-1}\right)-\frac{2}{9}\log|x+2|+C$

$=\frac{2}{9}\log|x-1|-\frac{1}{3(x-1)}-\frac{2}{9}\log|x+2|+C$

(ii) $\int \frac{3x-2}{(x+1)^2(x+3)}dx$

Sol :

Let $I=\frac{3x-2}{(x+1)^2(x+3)}dx$

$=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x+3)}$

$\frac{3x-2}{(x+1)^2(x+3)}=\frac{A(x+1)(x+3)+B(x+3)+C(x+1)^2}{(x+1)^2(x+3)}$

3x-2=A(x+1)(x+3)+B(x+3)+C(x+1)²

Putting x=-3 We get 

3(-3)-2=0+0+C(-3+1)²

-11=4C

$C=\frac{-11}{4}$

Putting x=-1 We get 3(-1)-2=0+B(-1+3)+0

-5=2B

$B=\frac{-5}{2}$

Putting x=2 We get

3×2-2=A(2+1)(2+3)+$\left(\frac{-5}{2} \right)(2+3)+\left(-\frac{11}{4}\right)\times (2+1)^2$

$4=15A-\frac{25}{2}-\frac{99}{4}$

$15A=\frac{4}{1}+\frac{25}{2}+\frac{99}{4}$

$15A=\frac{16+30+99}{4}=\frac{165}{4}$

∴$A=\frac{165}{4}\times \frac{1}{15}=\frac{11}{4}$

Now $I=\int \frac{3x-2}{(x+1)^2}(x+3)$

$=\int \left(\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{x+3}\right)dx$

$=\frac{11}{4}\log|x+1|-\frac{5}{2}\times \left(-\frac{1}{x+1}\right)-\frac{11}{4}\log|x+3|+C$

$=\frac{111}{4}\left[\log|x+1|-\log|x+3|\right]+\frac{5}{2(x+1)}+C$

$I=\frac{11}{4}\log\left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C$

(iii) $\int \frac{2x}{(x^2+1)(x^2+3)}dx$

Sol :

Let x²=z then 2xdx=dz

Now $I=\int \frac{dz}{(z+1)(z+3)}$

Let $\frac{1}{(z+1)(z+3)}=\frac{A}{z+1}+\frac{B}{z+3}$

$\frac{1}{(z+1)(z+3)}=\frac{A(x+3)+B(z+1)}{(z+1)(z+3)}$

1=A(z+3)+B(z+1)

Putting z=-1 We get

1A(-1+3)=0

∴$A=\frac{1}{2}$

Putting z=-3 We get 

1=0+B(-3+1)

∴$B=-\frac{1}{2}$

Now $I=\int \frac{1}{(z+1)(z+3)}$

$=\int \left(\frac{A}{z+1}+\frac{B}{z+3}\right)dz$

$=A\int \frac{1}{z+1}dz+B\int \frac{1}{z+3}dz$

$=\frac{1}{2}\left[\log|z+1|-\frac{1}{2}\log|z+3| \right]+C$

$=\frac{1}{2}\log\left|\frac{z+1}{z+3}\right|+C$

$=\frac{1}{2}\log\left|\frac{x^2+1}{x^2+3}\right|+C$

Question 13

$\int \frac{(3\sin \phi -2)\cos \phi d\phi}{5-\cos ^2 \phi -4\sin \phi}$

Sol :

Given : $I=\int \frac{(3\sin \phi -2)\cos \phi d\phi}{5-\cos ^2 \phi -4\sin \phi}$

$I=\int \frac{(3\sin \phi -2)\cos \phi d\phi}{5-1+\sin ^2 \phi -4\sin \phi}$

$=\int \frac{(3\sin \phi-2)\cos \phi d\phi}{4+\sin^2 \phi -4\sin \phi}$

Let sinф=z then cosфdф=dz

Now $I=\int \frac{(3\sin \phi -2)\cos \phi d\phi}{4+\sin ^2\phi -4\sin \phi}$

$=\int \frac{(3z-2)dz}{4+z^2-4z}$

$=\int \frac{(3z-2)dz}{(2-z)^2}$

Let $\frac{3z-2}{(2-z)^2}=\frac{A}{2-z}+\frac{B}{(2-z)^2}$

$\frac{3z-2}{(2-z)^2}=\frac{A(2-z)+B}{(2-z)^2}$

3z-2=A(2-z)+B

Putting z=2 We get 

3×2-2=A(2-2)+B

4=B

B=4

Putting z=1 We get

3×1-2=A(2-1)+4

1=A+4

A=-3

Now $I=\int \frac{(3z-2)dz}{(2-z)^2}$

$=\int \left(\frac{A}{2-z}+\frac{B}{(2-z)^2}\right)dz$

$=A\int \frac{1}{2-z}dz+B\int \frac{1}{(2-z)^2}dz$

$=\frac{-3\log|2-z|}{-1}+\dfrac{4\times \frac{-1}{2-z}+C}{-1}$

$=3\log|2-z|+\frac{4}{2-z}+C$

$I=3\log|2-\sin \phi |+\frac{4}{2-\sin \phi}+C$

Question 14

$\int \frac{3x+1}{(x-2)^2(x+2)}dx$

Sol :

Given : $I=\int \frac{3x+1}{(x-2)^2(x+2)}dx$

Let $\frac{3x+1}{(x-2)^2(x+2)}=\frac{A}{(x-2)}+\frac{B}{(x-2)^2}+\frac{C}{(x+2)}$

$\frac{3x+1}{(x-2)^2(x+2)}=\frac{A(x-2)(x+2)+B(x+2)+C(x-2)^2}{(x-2)^2(x+2)}$

3x+1=A(x-2)(x+2)+B(x+2)+C(x-2)²

Putting x=-2 We get

3(-2)+1=0+0+C(-2-2)²

-5=16C

$C=-\frac{5}{16}$

Putting x=2 We get 

3×2+1=0+B(2+2)+0

7=4B

$B=\frac{7}{4}$

Putting x=3 , We get

3×3+1=A(3-2)(3+2)+$\frac{7}{4}(3+2)+C(3-2)^2$

$10=5A+\frac{35}{4}+\left(-\frac{5}{16}\right)$

$10=5A+\frac{140-5}{16}$

$10=5A+\frac{135}{16}$

∴$5A=10-\frac{135}{16}=\frac{160-135}{16}=\frac{25}{16}$

∴$A=\frac{25}{16\times 5}=\frac{5}{16}$

Now $I=\int \frac{3x+1}{(x-2)^2(x+2)}dx$

$=\int \left(\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{x+2}\right)dx$

$=A\int \frac{1}{x-2}dx+B\int \frac{1}{(x-2)^2}dx+C\int \frac{1}{x+2}dx$

$=\frac{5}{16}\log|x-2|+\frac{7}{4}\times \left(\frac{-1}{(x-2)}\right)-\frac{5}{6}\log|x+2|+C$

$=\frac{5}{16}\left[\log|x-2|-\log|x+2|\right]-\frac{7}{4(x-2)}+C$

$I=\frac{5}{16}\log\left|\frac{x-2}{x+2}\right|-\frac{7}{4(x-2)}+C$

Question 15

$\int \frac{x^2+x+1}{(x-1)^3}dx$

Sol :

Let $\frac{x^2+x+1}{(x-1)^3}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}$

$\frac{x^2+x+1}{(x-1)^3}=\frac{A(x-1)^2+B(x-1)+C}{(x-1)^3}$

x²+x+1=A(x-1)²+B(x-1)+C

Putting x=1 We get 

1²+1+1=0+0+C

C=3

Putting x=0 We get

0+0+1=A×1-B+3

∴A-B=-2..(i)

Putting x=2 We get

4+2+1=A(2-1)²+B(2-1)+3

7=A+B+3

∴A+B=4...(ii)

Adding eq-(i) and (ii) We get

A-B+A+B=4-2

2A=2

∴A=1

Putting values of A in eq-(i) We get

A-B=-2

1-B=-2

∴B=3

Now $I=\int \frac{x^2+x+1}{(x-1)^3}dx$

$=\int \left(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}\right)dx$

$=A\int \frac{1}{x-1}dx+B\int \frac{1}{(x-1)^2}dx+C\int \frac{1}{(x-1)^3}dx$

$=1\times \log|x-1|+3\times \left(\frac{-1}{x-1}\right)+3\times \frac{1}{-2(x-1)^2}+C$

$=\log|x-1|-\frac{3}{x-1}-\frac{3}{2(x-1)^2}+C$

Question 16

$\int \frac{2}{(1+x)(1+x^2)}dx$

Sol :

Let $\frac{2}{(1-x)(1+x)^2}=\frac{A}{1-x}+\frac{Bx+C}{(1+x^2)}$

$\frac{2}{(1-x)(1+x^2)}=\frac{A(1+x^2)+(Bx+C)(1-x)}{(1-x)(1+x^2)}$

2=A(1+x²)+(Bx+C)(1-x)

Putting x=1 We get

2=A(1+1)+0=2A

∴A=1

Putting x=0 We get

2=A(1+0)+C×1

2=A+C

C=2-A=2-1

∴C=1

Putting x=-1 We get

2=A(1+1)+(B(-1)+1)(1+1)

2=2A(-B+1)×2

2=2×1+(-2B+2)

2=2+2-2B

-2=-2B

B=1

Now $\frac{2}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{B}{(1+x^2)}$

$=\frac{1}{(1-x)}+\frac{x+1}{(1+x^2)}$

∴$\int \frac{2}{(1-x)(1+x^2)}dx$ $=\int \frac{1}{1-x}dx+\int \frac{x+1}{1+x^2}dx$

$=-\log|1-x|+\int \frac{x}{1+x^2}dx+\int \frac{1}{1+x^2}dx$

$=-\log|1-x|+\frac{1}{2}\int \frac{2xdx}{1+x^2}+\tan^{-1}x+C$

$=-\log|1-x|+\frac{1}{2}\int \frac{dz}{z}+\tan^{-1}x+C$

$=-\log|1-x|+\frac{1}{2}\log z+\tan^{-1}x+C$

$I=-\log|1-x|+\frac{1}{2}\log|1+x^2|+\tan^{-1}x+C$

Question 17

$\int \frac{x}{(x^2+1)(x-1)}dx$

Sol :

Given : $I=\int \frac{x}{(x^2+1)(x-1)}dx$

$=\int \frac{x}{(x-1)(x^2+1)}dx$

Let $\frac{x}{(x-1)(x^2+1)}=\frac{A}{(x-1)}+\frac{Bx+C}{(x^2+1)}$

$\frac{x}{(x-1)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x-1)}{(x-1)(x^2+1)}$

x=A(x^2+1)+(Bx+C)(x-1)

Putting x=1 We get

1=A(1+1)+(B.1+C)(1-1)

1=2A+0

$A=\frac{1}{2}$

Putting x=0 We get

0=A.1+C(-1)⇒A-C=0

∴A=C$=\frac{1}{2}$

Putting x=-1 We get

-1=A(1+1)+(-B+C)(-1-1)

-1=2A+(C-B)(-2)

$-1=2\times \frac{1}{2}+2B-2C$

$-1=1+2B-2\times \frac{1}{2}=1+2B-1$

-1=2B

$B=-\frac{1}{2}$

Now $\frac{x}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{(x^2+1)}$

$=\frac{1}{2(x-1)}+\dfrac{-\frac{1}{2}x+\frac{1}{2}}{x^2+1}$

∴$\int \frac{x}{(x-1)(x^2+1)}=\frac{1}{2}\int \frac{1}{x-1}dx-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx$

$=\frac{1}{2}\log|x-1|-\frac{1}{4}\int \frac{2xdx}{x^2+1}+\frac{1}{2}\tan^{-1}x+C$

$I=\frac{1}{2}\log|x-1|-\frac{1}{4}\log|x^2+1|+\frac{1}{2}\tan^{-1}x+C$

Question 18

$\int \frac{x^2+x+1}{(x+2)(x^2+1)}dx$

Sol :

Let $\frac{x^2+x+1}{(x+2)(x^2+1)}=\frac{A}{(x+2)}+\frac{B}{(x^2+1)}$

$\frac{x^2+x+1}{(x+2)(x^2+1)}=\frac{A}{(x+2)}+\frac{Bx+C}{(x^2+1)}$

x²+x+1=A(x²+1)+(Bx+C)(x+2)

Putting x=-2 We get

4+(-2)+1=A(4+1)+0

2+1=5A

$A=\frac{3}{5}$

Putting x=0 We get

1=A(1)+C×2=A+2C

2C=1-A$=1-\frac{3}{5}=\frac{2}{5}$

∴$C=\frac{1}{5}$

Putting x=1 We get

$1+1+1=\frac{3}{5}(1+1)+(B.1+\frac{1}{5})(3)$

$3=\frac{6}{5}+3B+\frac{3}{5}=3B+\frac{9}{5}$

∴$3B=3-\frac{9}{5}=\frac{15-9}{5}=\frac{6}{5}$

∴$B=\frac{6}{5}\times \frac{1}{3}=\frac{2}{5}$

Now $\frac{x^2+x+1}{(x+2)(x^2+1)}=\frac{A}{(x+2)}+\frac{Bx+C}{(x^2+1)}$

$=\frac{3}{5}\times \frac{1}{(x+2)}+\dfrac{\frac{2}{5}x+\frac{1}{5}}{(x^2+1)}$

∴$\int \frac{x^2+x+1}{(x+2)(x^2+1)}=\int \frac{3}{5}\frac{1}{(x+2)}dx+\int \dfrac{\frac{1}{5}(2x+1)dx}{x^2+1}$

$I=\frac{3}{5}\log|x+2|+\int \frac{2xdx}{x^2+1}+\frac{1}{5}\int \frac{1}{x^2+1}dx$

$I=\frac{3}{5}\log|x+2|+\frac{1}{5}\log|x^2+1|+\frac{1}{5}\tan^{-1}x+C$

Question 19

$\int \frac{dx}{x^4+1}$

Sol :

Given : $I=\int \frac{dx}{x^4+1}=\int \frac{dx}{(x^2)^2-(1)^2}$

$I=\int \frac{dx}{(x-1)(x+1)(x^2+1)}$

Let $\frac{1}{(x-1)(x+1)(x^2+1)}=\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{C}{x^2+1}$

$\frac{1}{(x-1)(x+1)(x^2+1)}=\frac{A(x+1)(x^2+1)+B(x-1)(x^2+1)+C(x-1)(x+1)}{(x-1)(x+1)(x^2+1)}$

1=A(x+1)(x²+1)+B(x-1)(x²+1)+C(x-1)(x+1)

Putting x=1 we get

1=A(2)(2)+0+0

1=4A

∴$A=\frac{1}{4}$

Putting x=-1 we get

1=0+B(-2)(2)+0

1=-4B

∴$B=-\frac{1}{4}$

Putting x=2 we get

1=A(3)(5)+B(1)(5)+C(1)(3)

$1=\frac{15}{4}-\frac{15}{4}+3C=\frac{10}{4}+3C$

∴$3C=1-\frac{10}{4}$

$3C=\frac{4-10}{4}=\frac{-6}{4}$

∴$C=-\frac{6}{4}\times \frac{1}{3}=-\frac{1}{3}$

Now $I=\int \frac{1}{(x-1)(x+1)(x^2+1)}$

$=\int \left(\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x^2+1}\right)$

$=A\int  \frac{1}{x-1}dx+B\int \frac{1}{x+1}dx+C\int \frac{1}{x^2+1}dx$

$=\frac{1}{4}\log|x-1|-\frac{1}{4}\log|x+1|-\frac{1}{2}\tan^{-1}x+C$

$=\frac{1}{4}\left[\log|x-1|-\log|x+1|\right]-\frac{1}{2}\tan^{-1}x+C$

$I=\frac{1}{4}\log\left|\frac{x-1}{x+1}\right|-\frac{1}{2}\tan^{-1}x+C$

Question 20

$\int \frac{x^2+1}{x^2-5x+6}dx$

Sol :

Given : $I=\int \frac{x^2+1}{x^2-5x+6}dx$

$=\int \frac{x^2-4+5}{x^2-3x-2x+6}$

$=\int \frac{(x-2)(x+2)+5}{x(x-3)-2(x-3)}$

$=\int \frac{x+2}{x-3}dx+\int \frac{5}{(x-3)(x-2)}dx$

$=\int \frac{(x-3)+5}{(x-3)}dx+\int \frac{5}{(x-3)(x-2)}dx$

$=\int \frac{x-3}{x-3}dx+\int \frac{5}{x-3}dx+\int \frac{5}{(x-3)(x-2)}dx$

$=\int \frac{x-3}{x-3}dx+\int \frac{5}{x-3}dx+\int \frac{5}{(x-3)(x-2)}dx$

$=\int dx+5\int \frac{1}{x-3}dx+\int \frac{5}{(x-3)(x-2)}dx$

$=x+5\log|x-3|+\int \frac{5}{(x-3)(x-2)}dx$

Let $\frac{5}{(x-3)(x-2)}=\frac{A}{(x-3)}+\frac{B}{(x-2)}$

$\frac{5}{(x-3)(x-2)}=\frac{A(x-2)+B(x-3)}{(x-3)(x-2)}$

5=A(x-2)+B(x-3)

Putting x=3 we get

5=A(3-2)+0

∴A=5

Putting x=2 we get

5=0+B(2-3)

∴B=-5

Now $\frac{5}{(x-3)(x-2)}=\frac{5}{x-3}-\frac{5}{(x-2)}$

∴$\int \frac{5}{(x-3)(x-2)}dx$

$=\int \left( \frac{5}{(x-3)}-\frac{5}{(x-2)}\right)dx$

$=5\int \frac{1}{x-3}dx-5\int \frac{1}{x-2}dx$

=5log|x-3|-5log|x-2|

Now $I=x+5log|x-3|+\int \frac{5}{(x-3)(x-2)}dx$

=x+5log|x-3|+5log|x-3|-5log|x-2|+C

=x-5log|x-2|+10log|x-3+C

Question 21

$\int \frac{1-x^2}{x(1-2x)}dx$

Sol :

Given : $I=\int \frac{1-x^2}{x(1-2x)}dx$

$=\int \frac{1}{x(1-2x)}dx-\int \frac{x^2}{x(1-2x)}dx$

$=\int \frac{1}{x(1-2x)}dx-\int \frac{x}{(1-2x)}dx$

Let $\frac{1}{x(1-2x)}=\frac{A}{x}+\frac{B}{(1-2x)}$

$\frac{1}{x(1-2x)}=\frac{A(1-2x)+Bx}{x(1-2x)}$

1=A(1-2x)+Bx

Putting x=0 we get

1=A(1-0)+0

A=1

Putting $x=\frac{1}{2}$ we get

$1=A\left(1-2\times \frac{1}{2}\right)+\frac{B}{2}$

B=2

Now $\frac{1}{x(1-2x)}=\frac{A}{x}+\frac{B}{1-2x}$

$=\frac{1}{x}+\frac{2}{(1-2x)}$

∴$\int \frac{1}{x(1-2x)}=\int \left(\frac{1}{x}+\frac{2}{1-2x}\right)dx$

$=\int \frac{1}{x}dx+\int \frac{2}{1-2x}dx$

$=\log x+\int \frac{2dx}{1-2x}$

Let 1-2x=z then -2dx=dz

2dx=-dz

∴$\log x+\int \frac{-dz}{z}=\log x-\log z$

=logx-log|1-2x|

For $\int \frac{x}{1-2x}dx$

Let 1-2x=z then -2dx=dz $dx=-\frac{dz}{2}$

also 1-z=2x

∴$x=\frac{1-z}{2}$

Now $\int \frac{xdx}{1-2x}=\int \frac{(1-z)(-dz)}{2z2}$

$=-\frac{1}{4} \int \frac{(1-z)}{z}dz$

$=-\frac{1}{4} \int \left(\frac{1}{z}-1\right)dz$

$=-\frac{1}{4}\log z+\frac{1}{4}z$

$=-\frac{1}{4}\log(1-2x)+\frac{1-2x}{4}$

$=-\frac{1}{4}\log(1-2x)+\frac{1}{4}-\frac{2x}{4}$

∴$I=\int \frac{1}{x(1-2x)}dx-\int \frac{x}{1-2x}dx$

$=\log|x|-\log|1-2x|-\left[-\frac{1}{4}\log(1-2x)+\frac{1}{4}-\frac{x}{2}\right]$

$=\log|x|-\log|1-2x|+\frac{1}{4}\log(1-2x)-\frac{1}{4}+\frac{x}{2}+C$

$I=\frac{x}{2}+\log|x|-\frac{3}{4}\log|1-2x|+C$

Question 22

$\int \frac{x^3+x+1}{x^2-1}dx$

Sol :

Given : $I=\int \frac{x^3+x+1}{x^2-1}$

$=\int \frac{x^3}{x^2-1}dx+\int \frac{x}{x^2-1}dx+\int \frac{1}{x^2-1}dx$

$=\frac{1}{2}\int \frac{x^2.2xdx}{x^2-1}+\frac{1}{2}\int \frac{2xdx}{x^2-1}+\int \frac{1}{(x-1)(x+1)}dx$

For $\frac{1}{2}\int \frac{x^2.2xdx}{x^2-1}$

Let x²-1=z then 2xdx=dz

∴x²=z+1

Now $\frac{1}{2}\int \frac{x^2.2xdx}{x^2-1}=\frac{1}{2}\int \frac{(z+1)}{z}dz$

$=\frac{1}{2}\int \left(\frac{z}{z}+\frac{1}{z}\right)dx$

$=\frac{1}{2} \int 1dz+\frac{1}{2}\int \frac{1}{z}dz$

$=\frac{1}{2}z+\frac{1}{2}\log|z|$

$=\frac{1}{2}(x^2-1)+\frac{1}{2}\log(x^2-1)$

For $\int \frac{1}{(x-1)(x+1)}dx$

Let $\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}$

$=\frac{A(x+1)+B(x-1)}{(x-1)(x+1)}$

1=A(x+1)+B(x-1)

Putting x=1 we get 

1=2A+0

$A=\frac{1}{2}$

Putting x=-1 we get

1=0+B(-2)

$B=-\frac{1}{2}$

∴$\frac{1}{(x-1)(x+1)}=\frac{1}{2}\times \frac{1}{x-1}-\frac{1}{2}\times \frac{1}{x+1}$

∴$\int \frac{1}{(x-1)(x+1)}=\frac{1}{2}\int frac{1}{x-1}dx-\frac{1}{2}\int \frac{1}{x+1}dx$

$=\frac{1}{2}\log(x-1)-\frac{1}{2}\log(x+1)$

$=\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|$

Now $I=\frac{1}{2}\int \frac{x^2.2xdx}{x^2-1}+\frac{1}{2}\int \frac{2xdx}{x^2-1}+\int \frac{1}{(x-1)(x+1)}dx$

$=\frac{1}{2}(x^2-1)+\frac{1}{2}\log(x^2-1)+\frac{1}{2}\log(x^2-1)+\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|$

$=\frac{x^2}{2}-\frac{1}{2}+\log(x^2-1)+\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|$

$I=\frac{x^2}{2}+\log|x^2-1|+\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|$

$I=\frac{x^2}{2}\log|x^2-1|+\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|+C$

Now $I=\frac{1}{2}\int \frac{x^2.2xdx}{x^2-1}+\frac{1}{2}\int \frac{2xdx}{x^2-1}+\int \frac{1}{(x-1)(x+1)}dx$

$=\frac{1}{2}(x^2-1)+\frac{1}{2}\log(x^2-1)+\frac{1}{2}\log(x^2-1)+\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|$

$=\frac{x^2}{2}-\frac{1}{2}+\log(x^2-1)+\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|$

$I=\frac{x^2}{2}+\log|x^2-1|+\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|+C$

Question 23

$\int \frac{x^2}{x^2+6x+12}dx$

Sol :

Given : $I=\int \frac{x^2}{x^2+6x+12}dx$

$=\int \frac{x^2+6x+12-(6x+12)dx}{x^2+6x+12}$

$=\int \frac{x^2+6x+12}{x^2+6x+12}dx-\int \frac{6x+12+6-6}{x^2+6x+12}dx$

$=\int dx-\int \frac{6x+18-6}{x^2+6x+12}dx$

$=x-\int \frac{6x-18}{x^2+6x+12}dx+\int \frac{6}{x^2+6x+12}dx$

$=x-\int \frac{3(2x+6)dx}{x^2+6x+12}+6\int \frac{1}{x^2+2.x.3+9+3}dx$

$=x-3\int \frac{2x+6}{x^2+6x+12}dx+6\int \frac{1}{(x+3)^2+(\sqrt{3})^2}$

$=x-3\int \frac{2x+6}{x^2+6x+12}dx+6\int \frac{1}{(x+3)^2+(\sqrt{3})^2}$

$=x-3\log|x^2+6x+12|+6\times \frac{1}{\sqrt{3}}\tan^{-1}\frac{(x+3)}{\sqrt{3}}+C$

$=x-3\log|x^2+6x+12|+2\sqrt{3}\tan^{-1} \left(\frac{x+3}{\sqrt{3}}\right)+C$

Question 24

$\int \frac{1-\cos x}{\cos x(1+\cos x)}dx$

Sol :

Given : $I=\int \frac{1-\cos x}{\cos x(1+\cos x)}dx$

$=\int \frac{2\sin ^2 \frac{x}{2}dx}{2\cos ^2 \frac{x}{2}\left(\dfrac{1-\tan^{2}\frac{x}{2}}}{1+\tan^{2}\frac{x}{2}}\right)}$

$=\int \dfrac{\tan^{2}\frac{x}{2}.\sec^2\frac{x}{2}dx}{1-\tan^2\frac{x}{2}}$

Let $\tan \frac{x}{2}=z$ then $\frac{1}{2}\sec^2\frac{x}{2}=dz$ 

$=\sec^2 \frac{x}{2}=2dz$

Now $i=\int \frac{z^2 2dz}{1-z^2}$

$=2\int \frac{z^2}{1-z^2}dz$

$=-2\int \frac{z^2}{z^2-1}dz$

$=-2\int \frac{z^2-1+1}{z^2-1}dz$

$=-2\int \frac{z^2-1}{z^2-1}dz-2\int \frac{1}{z^2-1}dz$

$=-2\int dz+2\int \frac{1}{1-z^2}dz$

$=-27+2\int \frac{1}{(1-z)(1+z)}dz$

Let $\frac{1}{(1-z)(1+z}=\frac{A}{1-z}+\frac{B}{1+z}=\frac{A(1+z)+B(1-z)}{(1-z)(1+z)}$

∴=A(1+z)+B(1-z)

Putting z=1 we get 

1=A(2)+0

$A=\frac{1}{2}$

Putting z=-1 we get

1=0+B(2)

$B=\frac{1}{2}$

∴$\frac{1}{(1-z)(1+z)}=\frac{A}{1-z}+\frac{B}{1+z}=\frac{1}{2(1-z)}+\frac{1}{2(1+z)}$

Now $2\int \frac{1}{(1-z)(1+z)}=2\int \left(\frac{1}{2(1-z)+\frac{1}{2(1+z)}}\right)dz$

$=2\times \frac{1}{2} \int \frac{1}{1-z}dz+2\times \frac{1}{2}\int \frac{1}{1+z}dz$

=-log|1-z|+log|1+z|

$=\log\left|\frac{1+z}{1-z}\right|$

∴$I=-2z+2\int \frac{1}{(1-z)(1+z)}dz$

$=-2z+\log \left|\frac{1+z}{1-z}\right|+C$

$=\log\left|\frac{1+z}{1-z}\right|-2z+C$

$=\log\left|\dfrac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right|-2\tan \frac{x}{2}+C$

$I=\log|\sec x+\tan x|-2\tan \frac{x}{2}+C$

Question 25

$\int \frac{1+\sin xdx}{\sin x(1+\cos x)}$

Sol :

Given : $I=\int \frac{1+\sin xdx}{\sin x(1+\cos x)}$

$=\int \frac{1}{\sin x(1+\cos x)}dx+\int \frac{\sin x}{\sin x (1+\cos x)}dx$

$=\int \frac{1}{\sin x (1+\cos x)}dx+\int \frac{1}{1+\cos x}dx$

$=\int \dfrac{1}{\dfrac{2\tan \frac{x}{2}}{1+\tan ^2\frac{x}{2}}\left(2\cos ^2 \frac{x}{2}\right)}dx+\int \frac{dx}{2\cos ^2 \frac{x}{2}}$

$=\frac{1}{4}\int \frac{\sec^2 \frac{x}{2}.\sec^2\frac{x}{2}}{\tan \frac{x}{2}}+\frac{1}{2}\int \sec ^2 \frac{x}{2}dx$

$=\frac{1}{4}\int \dfrac{\sec^2 \frac{x}{2}.\sec^2 \frac{x}{2}}{\tan\frac{x}{2}}+\frac{1}{2}\int \sec^2 \frac{x}{2}dx$

$=\frac{1}{4}\int \dfrac{\left(1+\tan^2 \frac{x}{2}\right)}{\tan \frac{x}{2}}.\sec ^2 \frac{x}{2}dx+\frac{1}{2}\tan \frac{x}{2}\times 2+C$

$=\frac{1}{4}\int \dfrac{\left(1+\tan^2 \frac{x}{2}\right)}{\tan \frac{x}{2}}.\sec ^2 \frac{x}{2}dx+\tan \frac{x}{2}+C$

Let $\tan \frac{x}{2}=z$ then $\frac{1}{2}\sec^2 \frac{x}{2} dx=dz$

$\sec^2 \frac{x}{2}dx=2dz$

Now $i=\frac{1}{4}\int \left(\frac{1+z^2}{z}\right)2dz+\tan \frac{x}{2}+C$

$=\frac{1}{2}\int \left(\frac{1}{z}+z\right)dz+\tan \frac{x}{2}+C$

$=\frac{1}{2}\int \frac{1}{z}dz+\frac{1}{2}\int zdz+\tan \frac{x}{2}+C$

$=\frac{1}{2}\log|z|+\frac{1}{2}\frac{z^2}{2}+\tan \frac{x}{2}+C$

$=\frac{1}{2}\log \left|\tan \frac{x}{2}\right|+\frac{1}{4}\tan ^2\frac{x}{2}+\tan \frac{x}{2}+C$

$=\frac{1}{2}\log \left|\tan \frac{x}{2}\right|+\frac{1}{4}\left(\sec ^2\frac{x}{2}-1\right)+\tan \frac{x}{2}+C$

$=\frac{1}{2}\log \left|\tan \frac{x}{2}\right|+\frac{1}{4}\left(\sec ^2\frac{x}{2}-\frac{1}{4}\right)+\tan \frac{x}{2}+C$

$I=\frac{1}{2}\log \left|\tan \frac{x}{2}\right|+\frac{1}{4}\sec ^2 \frac{x}{2}+\tan \frac{x}{2}+C$

Question 26

$\int \frac{dx}{\sin x(3+2\cos x)}$

Sol :

Given : $I=\int \frac{dx}{\sin x(3+2\cos x)}$

$=\int \frac{\sin xdx}{\sin x.\sin x(3+2\cos x) }$

$=\int \frac{\sin xdx}{\sin ^2 x(3+2\cos x)}$

$=\int \frac{\sin xdx}{(1-\cos^2x)(3+2\cos x)}$

$=\int \frac{\sin xdx}{(1-\cos x)(1+\cos x)(3+2\cos x)}$

Let cosx=z then 

-sinxdx=dz 

sinxdx=-dz

Now $I=\int \frac{-dz}{(1-z)(1+z)(3+2z)}$

Let $\frac{1}{(1-z)(1+z)(3+2z)}=\frac{A}{(1-z)}+\frac{B}{(1+z)}+\frac{C}{(3+2z)}$

$\frac{1}{(1-z)(1+z)(3+2z)}=\frac{A(1+z)(3+2z)+B(1-z)(3+2z)+C(1-z)(1+z)}{(1-z)(1+z)(3+2z)}$

1=A(1+z)(3+2z)+B(1-z)(3+2z)+C(1-z)(1+z)

Putting z=1 We get 

1=A(1+1)(3+2×1)+0+0

1=A×2×5=10A

∴$A=\frac{1}{10}$

Putting z=1 We get 

1=0+B(1-(-1)(3-2)+0

1=2B

$B=\frac{1}{2}$

Putting $z=\frac{-3}{2}$ We get 

1=0+0+$C\left(1+\frac{3}{2}\right)\left(1-\frac{3}{2}\right)$

$1=C\times \frac{5}{2}\times \frac{-1}{2}$

$C=-\frac{4}{5}$

Now $I=-\int \frac{dz}{(1-z)(1+z)(3+2z)}$

$=-\int \left(\frac{A}{1-z}+\frac{B}{1+z}+\frac{C}{3+2z}\right)$

$=-A\int \frac{1}{1-z}dz-B\int \frac{1}{1+z}dz-C\int \frac{1}{3+2z}dz$

$=-\frac{1}{10}\frac{\log|1-z|}{-1}-\frac{1}{2}\log|1+z|-\left(-\frac{4}{5}\right)\frac{\log |3+2z|}{2}$

$=\frac{1}{10}\log|1-\cos x|-\frac{1}{2}\log |1+\cos x|+\frac{2}{5}\log|3+2\cos x|+C$

Question 27

$\int \frac{\sin x}{\sin 4x}dx$

Sol :

Given : $I=\int \frac{\sin x }{\sin 4x}dx$

$=\int \frac{\sin x}{\sin 2x}dx$

$=\int \frac{\sin x}{2\sin 2x.cos 2x}dx$

$=\int \frac{\sin x}{2.2\sin x.\cos x(1-2\sin ^2 x)}dx$

$=\frac{1}{4}\int \frac{dx}{\cos x(1-2\sin ^2x)}$

$=\frac{1}{4}\int \frac{\cos xdx}{\cos x.\cos x(1-2\sin^2 x)}$

$=\frac{1}{4}\int \frac{\cos xdx}{\cos^2 x(1-2\sin^2 x)}$

$=\frac{1}{4}\int \frac{\cos xdx}{(1-\sin ^2x )(1-2\sin^2 x)}$

Let sinx=z then cosxdx=dz

Now $I=\frac{1}{4} \int \frac{dz}{(1-z^2)(1-2z^2)}$

$=\frac{1}{4}\int \frac{dz}{(1-z)(1+z)(1-2z^2)}$

Let $\frac{1}{(1-z)(1+z)(1-2z^2)}=\frac{A}{(1-z)}+\frac{B}{(1+z)}+\frac{C}{(1-2z^2)}$

$\frac{1}{(1-z)(1+z)(1-2z^2)}=\frac{A(1+z)(1-2z^2)+B(1-z)(1-2z^2)+C(1-z)(1+z)}{(1-z)(1+z)(1-2z^2)}$

1=A(1+z)(1-2z²)+B(1-z)(1-2z²)+C(1-z)(1+z)

Putting z=1 we get

1=A(2)(-1)=-2A

∴$A=-\frac{1}{2}$

Putting z=-1 we get

1=0+B(2)(-1)=-2B

∴$B=-\frac{1}{2}$

Putting $z=\frac{1}{\sqrt{2}}$ we get

$1=0+0+C\left(1-\frac{1}{\sqrt{2}}\right)\left(1+\frac{1}{\sqrt{2}}\right)$

$1=c\left(1-\frac{1}{2}\right)=\frac{C}{2}$

∴C=2

Now $I=\frac{1}{4}\int \frac{1}{(1-z)(1+z)(1-2z^2)}$

$=\frac{1}{4}\int \left(\frac{A}{1-z}+\frac{B}{1+z}+\frac{C}{1-2z^2}\right)dz$

$=\frac{1}{4}A\int \frac{1}{1+-z}dz+\frac{1}{4}B\int \frac{1}{1+z}dz+\frac{1}{4}C\int \frac{1}{1-2z^2}dz$

$=\frac{1}{4}\times \frac{-1}{2}\frac{\log|1-z|}{-1}+\frac{1}{4}\times \frac{-1}{2}\log|1+z|+\frac{2}{4}\int \frac{1}{(1)^2-(\sqrt{2}z)^2}$

$=\frac{1}{8}\log|1-z|-\frac{1}{8}\log|1+z|+\frac{1}{2}\times \frac{1}{2\sqrt{2}}\log\left|\frac{1+\sqrt{2}z}{1-\sqrt{2}z}\right|$

$=\frac{1}{8}\left[\log|1-z|-\log|1+z|+ \frac{1}{4\sqrt{2}}\log\left|\frac{1+\sqrt{2}z}{1-\sqrt{2}z}\right|\right]+C$

$=-\frac{1}{8}\log\left|\frac{1+z}{1-z}\right|+\frac{1}{4\sqrt{2}}\log\left|\frac{1+\sqrt{2}z}{1-\sqrt{2}z}\right|+C$

$I=\frac{-1}{8}\log\left|\frac{1+\sin x}{1-\sin x}\right|+\frac{1}{4\sqrt{2}}\log\left|\frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x}\right|+C$

Question 28

$\int \frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}dx$

Sol :

Dividing (x-1)(x-2)(x-3) by (x+1)(x+2)(x+3) we get quotient 1

Let $\frac{(x-1)(x-2)(x-3)}{(x-1)(x+)(x+3)}=1+\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x+3)}$

(x-1)(x-2)(x-3)=(x+1)(x+2)(x+3)+A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)

Putting x=-1 we get

(-2)(-3)(-4)=A(1)(2)+0+0

∴2A=-24

A=-12

Putting x=-2 we get

(-3)(-4)(-5)=0+B(-1)(1)=-B

∴-B=-60

B=60

Putting x=-3 we get

(-4)(-5)(-6)=0+0+C(-2)(-1)

∴2C=-120

C=-60

Now $I=\int \frac{(x-1)(x-2)(x-3)dx}{(x+1)(x+)(x+3)}$

$=\int \left(1+\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}\right)dx$

$=\int dx+A\int \frac{1}{x+1}dx+B\int \frac{1}{x+2}dx+C\int \frac{1}{x+3}dx$

=x-12log(x+1)+60log(x+2)-60log(x+3)+C

=x+60log(x+2)-12log(x+1)-60log(x+)+C

=x+12×5log(x+2)-12log(x+1)-12×5log(x+3)+C

 =x+12log(x+2)⁵-12log(x+1)-12log(x+3)⁵+C

 =x+12log(x+2)⁵-12[log(x+1)+log(x+3)⁵]+C

=x+12log(x+2)⁵-12log(x+1)(x+3)⁵+C

=x+12[log(x+2)⁵-log(x+1)(x+3)⁵]+C

$I=x+12\log\left|\frac{(x+2)^5}{(x+1)(x+3)^5}\right|+C$

Question 29

$\int \frac{dx}{(x+1)^2(x^2+1)}$

Sol :

Let $\frac{1}{(x+1)^2(x^2+1)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{Cx+D}{(x^2+1)}$

$\frac{1}{(x+1)^2(x^2+1)}=\frac{A(x+1)(x^2+1)+B(x^2+1)+(Cx+D)(x+1)^2}{(x+1)^2(x^2+1)}$

1=A(x+1)(x²+1)+B(x²+1)+(Cx+D)(x+1)²

Putting x=-1 we get

1=0+B(2)+0

$B=\frac{1}{2}$

Putting x=0 we get

1=A(1)(1)+B+D(1)

∴A+B+D=1

∴A+B=1-B$=1-\frac{1}{2}$

∴$A+D=\frac{1}{2}$..(i)

∴$A=\frac{1}{2}-D=\frac{1}{2}-0=\frac{1}{2}$

Putting x=1 we get 

1=A(2)(2)+B(2)+(C+D)4

1=4A+4C+4D+$2\times \frac{1}{2}$

1=4(A+D+C)=0

A+D+C=0

$\frac{1}{2}+C=0$

$C=-\frac{1}{2}$

Putting x=2 we get

1=A(3)()+B(5)+(2C+D)9

1=15A+5B+18C+9D

$15A+5B+18\times \left(-\frac{1}{2}\right)+9D=1$

$15A+\frac{5}{2}-9+9D=1$

$15A+9D=10-\frac{5}{2}=\frac{15}{2}$

$15\left(\frac{1}{2}-D\right)+9D=\frac{15}{2}$

$\frac{15}{2}-15D+9D=\frac{15}{2}$

-6D=0

D=0

Now $I=\int \frac{dx}{(x+1)^2(x^2+1)}$

$=\int \left(\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{Cx+D}{x^2+1}\right)dx$

$=A\int \frac{1}{x+1}dx+B\int \frac{1}{(x+1)^2}dx+C\int \frac{x}{x^2+1}dx$

$=\frac{1}{2}\log|x+1|+\frac{1}{2}\times \frac{-1}{(x+1)}-\frac{1}{2\times 2}\int \frac{2xdx}{x^2+1}$

$I=\frac{1}{2}\log|x+1|-\frac{1}{2(x+1)}-\frac{1}{4}\log(x^2+1)+C$

Question 30

$\int \frac{x^2}{(x^2+1)(x^2+4)}dx$

Sol :

Given : $I=\int \frac{x^2}{(x^2+1)(x^2+4)}dx$

Let $\frac{y}{(y+1)(y+4)}=\frac{A}{(y+1)}+\frac{B}{(y+4)}$

$\frac{y}{(y+1)(y+4)}=\frac{A(y+4)+B(y+1)}{(y+1)(y+4)}$

y=A(y+4)+B(y+1)

Putting y=-1 we get

-1=A(-1+4)+0=3A

∴$A=-\frac{1}{3}$

Putting y=-4 we get

-4=0+B(-4+1)⇒-3B

∴$B=\frac{4}{3}$

Now $I=\int \frac{y}{(y+1)(y+4)}dy$

$=\int \frac{A}{y+1}dy+\int \frac{B}{y+4}dy$

$=A\int \frac{1}{y+1}dy+B\int \frac{1}{y+4}dy$

$=-\frac{1}{3}\int \frac{1}{x^2+1}dx+\frac{4}{3}\int \frac{1}{x^2+(2)^2}dx$

$=-\frac{1}{3}\tan^{-1}x+\frac{4}{3}\times \frac{1}{2}\tan^{-1}\frac{x}{2}+C$

$I=-\frac{1}{3}\tan^{-1}x+\frac{2}{3}\tan^{-1}\frac{x}{2}+C$

Question 31

$\int \frac{(x^2+1)}{(x^2+4)(x^2+25)}dx$

Sol :

Given : $I=\int \frac{(x^2+1)}{(x^2+4)(x^2+25)}dx$

Let x²=y then $I=\int \frac{(y+1)}{(y+4)(y+25)}dx$

Let $\frac{y+1}{(y+4)(y+25)}=\frac{A}{(y+4)}+\frac{B}{(y+25)}$

$\frac{y+1}{(y+4)(y+25)}=\frac{A(y+25)+B(y+4)}{(y+4)(y+25)}$

y+1=A(y+25)+B(y+4)

Putting y=-4 we get 

-4+1=A(-4+25)+0

-3=21A

$A=\frac{-3}{21}=-\frac{1}{7}$

Putting y=-4 we get 

-4+1=A(-4+25)+0

-3=21A

$A=\frac{-3}{21}=-\frac{1}{7}$

Putting y=-25 we get 

-25+1=0+B(-25+4)

-21B=-24

$B=\frac{24}{21}=\frac{8}{7}$

Now $I=\int \frac{y+1}{(y+4)(y+25)}dx$

$=\int \left(\frac{A}{y+4}+\frac{B}{y+25}\right)dx$

$=A\int \frac{1}{y+4}dx+B\int \frac{1}{y+25}dx$

$=-\frac{1}{7}\int \frac{1}{x^2+4}dx+\frac{8}{7}\int \frac{1}{x^2+25}dx$

$\left[\int \frac{1}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}\right]$

$=-\frac{1}{7}\int \frac{1}{(x)^2+(2)^2}dx+\frac{8}{7}\int \frac{1}{(x)^2+(5)^2}dx$

$I=\frac{-1}{7}\times \frac{1}{2}\tan^{-1}\frac{x}{2}+\frac{8}{7}\times \frac{1}{5}\tan^{-1}\frac{x}{5}+C$

$I=-\frac{1}{14}\tan^{-1}\frac{x}{2}+\frac{8}{35}\tan^{-1}\frac{x}{5}+C$