Exercise 19.13
Example-3
$\int \frac{dx}{3+2x-x^2}$
Sol :
Given : $I=\int \frac{dx}{3+2x-x^2}$
$=\int \frac{dx}{3-(x^2-2x+1-1)}$
$=\int \frac{dx}{3-(x^2-2x+1)+1}$
$=\int \frac{dx}{4-(x^2-2x+1)}$
$=\int \frac{dx}{(2)^2-(x-1)^2}$
$=\frac{1}{2\times 2}\log \left|\frac{2+x-1}{2-x+1}\right|+C$
$\left[\therefore \int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|\right]$
$=\frac{1}{4}\log \left|\frac{x+1}{3-x}\right|+C$
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: $I=\int \frac{x}{(x+1)(x+2)}dx$
$=\frac{x}{(x+1)(x+2)}=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}$
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 :
$I=\int \frac{\cos x}{(1+\sin x)(2+\sin x)}dx$
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=xex also $x=\frac{z}{e^x}$ and
then dz=(xex+ex1)dx
dz=ex(x+1)dx
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 ex=z then exdx=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 ex+log|ex-1|+C
(∴ logex=x)
=-x+log|ex-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=ex then exdx=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+ex|-log|2+ex|+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 ex=z then exdx=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 ex+log|1+ex|-2log|1+2ex|+C
I=x+log|1+ex|-2log|1+2ex|+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 : $I=\int \frac{(3\sin \phi -2)\cos \phi d\phi}{5-\cos ^2 \phi -4\sin \phi}$
$=\int \frac{(3\sin \phi-2)\cos \phi d\phi}{4+\sin^2 \phi -4\sin \phi}$
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 : $I=\int \frac{x^2+1}{x^2-5x+6}dx$
$=\int \frac{(x-2)(x+2)+5}{x(x-3)-2(x-3)}$
$=\int \frac{x+2}{x-3}dx+\int \frac{5}{(x-3)(x-2)}dx$
$=\int \frac{(x-3)+5}{(x-3)}dx+\int \frac{5}{(x-3)(x-2)}dx$
$=\int \frac{x-3}{x-3}dx+\int \frac{5}{x-3}dx+\int \frac{5}{(x-3)(x-2)}dx$
$=\int \frac{x-3}{x-3}dx+\int \frac{5}{x-3}dx+\int \frac{5}{(x-3)(x-2)}dx$
$=\int dx+5\int \frac{1}{x-3}dx+\int \frac{5}{(x-3)(x-2)}dx$
$=x+5\log|x-3|+\int \frac{5}{(x-3)(x-2)}dx$
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 : $I=\int \frac{x^2}{x^2+6x+12}dx$
$=\int \frac{x^2+6x+12}{x^2+6x+12}dx-\int \frac{6x+12+6-6}{x^2+6x+12}dx$
$=\int dx-\int \frac{6x+18-6}{x^2+6x+12}dx$
$=x-\int \frac{6x-18}{x^2+6x+12}dx+\int \frac{6}{x^2+6x+12}dx$
$=x-\int \frac{3(2x+6)dx}{x^2+6x+12}+6\int \frac{1}{x^2+2.x.3+9+3}dx$
$=x-3\int \frac{2x+6}{x^2+6x+12}dx+6\int \frac{1}{(x+3)^2+(\sqrt{3})^2}$
$=x-3\int \frac{2x+6}{x^2+6x+12}dx+6\int \frac{1}{(x+3)^2+(\sqrt{3})^2}$
$=x-3\log|x^2+6x+12|+6\times \frac{1}{\sqrt{3}}\tan^{-1}\frac{(x+3)}{\sqrt{3}}+C$
$=x-3\log|x^2+6x+12|+2\sqrt{3}\tan^{-1} \left(\frac{x+3}{\sqrt{3}}\right)+C$
Question 24
$\int \frac{1-\cos x}{\cos x(1+\cos x)}dx$
Sol :
Given : $I=\int \frac{1-\cos x}{\cos x(1+\cos x)}dx$
$=\int \dfrac{\tan^{2}\frac{x}{2}.\sec^2\frac{x}{2}dx}{1-\tan^2\frac{x}{2}}$
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 $I=\int \frac{y+1}{(y+4)(y+25)}dx$
$=\int \left(\frac{A}{y+4}+\frac{B}{y+25}\right)dx$
$=A\int \frac{1}{y+4}dx+B\int \frac{1}{y+25}dx$
$=-\frac{1}{7}\int \frac{1}{x^2+4}dx+\frac{8}{7}\int \frac{1}{x^2+25}dx$
$\left[\int \frac{1}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}\right]$
$=-\frac{1}{7}\int \frac{1}{(x)^2+(2)^2}dx+\frac{8}{7}\int \frac{1}{(x)^2+(5)^2}dx$
$I=\frac{-1}{7}\times \frac{1}{2}\tan^{-1}\frac{x}{2}+\frac{8}{7}\times \frac{1}{5}\tan^{-1}\frac{x}{5}+C$
$I=-\frac{1}{14}\tan^{-1}\frac{x}{2}+\frac{8}{35}\tan^{-1}\frac{x}{5}+C$
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