KC Sinha Solution Class 12 Chapter 19 Indefinite Integrals Exercise 19.3

Exercise 19.3

Question 1
$\displaystyle\int(x+2) \sqrt{3 x+5} d x$
Sol :
Let $z=3 x+5$ then $\dfrac{d z}{d x}=3$

∴$d x=\dfrac{d z}{3}$

Again z=3 x+5 ∴ $x=\dfrac{2-5}{3}$

Now, $\displaystyle\int(x+2) \cdot \sqrt{3 x+5} d x$

$=\displaystyle\int\left(\dfrac{z-5}{3}+2\right) \cdot \sqrt{z}  \dfrac{dz}{3}$

$=\frac{1}{3} \displaystyle\int\left(\frac{z-5+6}{3}\right) \sqrt{z} d z$

$=\frac{1}{9} \displaystyle\int(z+1) \sqrt{z} d z$

$=\dfrac{1}{9} \displaystyle\int z^{\tfrac{3}{2}} dz+\dfrac{1}{9} \displaystyle\int z^{\tfrac{1}{2}} d z$

$=\dfrac{1}{9} \cdot \dfrac{2}{5} \cdot z^{\dfrac{5}{2}}+\dfrac{1}{9} \cdot \dfrac{2}{3} \cdot z^{\tfrac{3}{2}}+c$

=$z^{\tfrac{3}{2}}\left[\dfrac{2}{45}z+\dfrac{2}{27}\right]$+c

=$z^{\tfrac{3}{2}}\left[\dfrac{6z+10}{135}\right]$+c

=$\dfrac{1}{135}.(3x+5)^{\tfrac{3}{5}}.\left[6(3x+5)+10\right]$+c

=$\dfrac{1}{135}.(3x+5)^{\tfrac{3}{5}}.\left[18x+40\right]$+c

=$\dfrac{2}{135}.(9x+20).(3x+5)^{\tfrac{3}{2}}$+c


Question 2
$\displaystyle\int (x-1).\sqrt{x+1}dx$
Sol :
Let z=x+1 then $\dfrac{dz}{dx}=1$

∴dx=dz

Again , z=x+1 ∴x=z-1

Now, $\displaystyle\int (x-1)\sqrt{x+1}dx$

=$\displaystyle\int (z-1-1).\sqrt{z}dz$

=$\displaystyle\int (z-2).\sqrt{z}dz$

=$\displaystyle\int z^{\tfrac{3}{5}}dz-2\displaystyle\int z^{\tfrac{1}{2}}dz$

=$\dfrac{2}{5}.z^{\tfrac{5}{2}}-2.\dfrac{2}{3}.z^{\tfrac{3}{2}}$+c

=$\dfrac{2}{5}.z^{\tfrac{5}{2}}-\dfrac{4}{3}.z^{\tfrac{3}{2}}$+c

=$z^{\tfrac{3}{2}}\left[\dfrac{2}{5}z-\dfrac{4}{3}\right]$+c

=$z^{\tfrac{3}{2}}\left[\dfrac{6z-20}{15}\right]$+c

=$z^{\tfrac{3}{2}}\left[\dfrac{6(x+1)-20}{15}\right]$+c

=$(x+1)^{\tfrac{3}{2}}.\left(\dfrac{6x+6-20}{15}\right)$+c

=$(x+1)^{\tfrac{3}{2}}.2\left(\dfrac{3x-7}{15}\right)$+c

=$\dfrac{2}{15}.(x+1)^{\tfrac{3}{2}}.(3x-7)$+c

Question 3
$\int(x+2) \cdot \sqrt{2 x+1} d x$
Sol:
Let $z=2 x+1$ then $\frac{d z}{d x}=2 \therefore d x=\frac{d z}{2}$

again, z=2 x+1 $\therefore x=\frac{z-1}{2}$

Now, $\left.\int\left(\frac{z-1}{2}\right)+2\right) \cdot \sqrt{z} \frac{d z}{2}=\int\left(\frac{z-1+4}{2}\right) \sqrt{z} \frac{d z}{2}$

$=\frac{1}{4} \int(z+3) \cdot \sqrt{z} d z$

$=\frac{1}{4} \int z^{\frac{3}{2}} d z+\frac{3}{4} \int z^{\frac{1}{2}} d z$


$=\frac{1}{4} \cdot \frac{2}{5} \cdot z^{\frac{5}{2}}+\frac{3}{4} \cdot \left(\dfrac{z}{\frac{3}{2}}\right)^{\frac{3}{2}}+c$

$=\dfrac{1}{10}z^{\frac{5}{2}}+\dfrac{1}{2}z^{\frac{3}{2}}+c$

=$z^{\frac{3}{2}}\left[\dfrac{1}{10}z+\dfrac{1}{2}\right]+c$

=$(2x+1)^{\frac{3}{2}}\left[\dfrac{1}{10}(2x+1)+\dfrac{1}{2}\right]+c$

=$(2x+1)^{\frac{3}{2}}\left[\dfrac{2x+1+5}{10}\right]+c$

=$(2x+1)^{\frac{3}{2}}\left[\dfrac{2x+6}{10}\right]+c$

=$\dfrac{2}{10}(x+3)(2x+1)^{\frac{3}{2}}+c$


Question 4
$\int x \cdot \sqrt{2 x+1} d x$
Sol :
Let z=2x+1 , then $\frac{d z}{d x}=2$ $\therefore d x=\frac{d z}{2}$

Again Z=2x+1 $\therefore x=\frac{z-1}{2}$

Now , $\int x \cdot \sqrt{2 x+1} d x$

$=\int \frac{z-1}{2} \cdot \sqrt{z} \frac{d z}{2}$

$=\frac{1}{4} \int z^{\frac{3}{2}}-z^{\frac{1}{2}} d z$

$=\frac{1}{4} \int z^{\frac{3}{2}} d z-\frac{1}{4} \int z^{\frac{1}{2}} d z$

$=\frac{1}{4} \int z^{\frac{3}{2}}-z^{\frac{1}{2}} d z$

$=\frac{1}{4} \int z^{\frac{3}{2}} d z-\frac{1}{4} \int z^{\frac{1}{2}} d z$

$=\frac{1}{4} \cdot \frac{2}{5} \cdot z^{\frac{5}{2}}-\frac{1}{4} \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}+c$

$=\frac{1}{10} z^{\frac{5}{2}}-\frac{1}{6} z^{\frac{3}{2}}+c$

$=z^{\frac{3}{2}}\left[\frac{1}{10} z-\frac{1}{6}\right]+c$

$=z^{\frac{3}{2}}\left[\frac{2 x+1}{10}-\frac{1}{6}\right]+c$

$=(2 x+1)^{\frac{3}{2}} \cdot\left(\frac{6 x-2}{30}\right)+c$

$=\frac{2(3 x-1)}{30} \cdot(2 x+1)^{\frac{3}{2}}+c$

$=\frac{1}{15}(3 x-1) \cdot(2 x+1)^{\frac{3}{2}}+c$

Question 5
$\int x \cdot(3 x-5)^{4} d x$
Sol :
Let z=3x-5 then $\frac{d z}{d x}=3 \therefore d x=\frac{d z}{3}$

Again z=3x-5 $\therefore x=\dfrac{z+5}{3}$

Now $\int x \cdot(3 x-5)^{4} d x$ $=\int \frac{z+5}{3} \cdot z^{4} \frac{d z}{3}$

$=\frac{1}{9} \int z^{5} d z+\frac{5}{9} \int z^{4} d z$

$=\frac{1}{9} \cdot \frac{z^{6}}{6}+\frac{5}{9} \cdot \frac{z^{5}}{5}+c$

$=z^{5}\left[\frac{2}{5}+\frac{1}{5}\right]+c$

$=z^{5}\left[\frac{z+6}{54}\right]+c$

$=\frac{1}{54} \cdot(3 x-5)^{5} \cdot(3 x+1)+c$


Question 11
$\int x^{2} \cdot \sqrt{x+2} d x$
Sol :
Let z=x+2 \therefore \frac{d z}{d x}=1 \quad \therefore \quad d z=d x
Again z=x+2 , x=z-2

Now $\int x^{2} \cdot \sqrt{x+2} d x=\int(z-2)^{2} \cdot \sqrt{z} d z$

$=\int\left(z^{2}+4-4 z\right) \sqrt{z} d z$

$=\int z^{\frac{1}{2}} d z+4 \int z^{\frac{1}{2}} d z-4 \int z^{\frac{3}{2}} d z$

$=\frac{2}{7} \cdot 2^{\frac{7}{2}}+4 \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}-4 \cdot \frac{2}{5} \cdot z^{\frac{5}{2}}+c$

$=2 z^{\frac{3}{2}}\left[\frac{1}{7} z^{2}+\frac{4}{3}-\frac{4}{5} z\right]+c$

=2(x+2)^{\frac{3}{2}}\left(\frac{(x+2)^{2}}{7}-\frac{4}{5}(x+2)+\frac{4}{3}\right]+c


Question 12
$\int\left(2 x^{2}+3\right) \sqrt{x+2} d x$
Sol :
Let z=x+2

$\therefore \frac{d z}{d x}=1 \quad \therefore \quad d z=d x$

again z=x+2 ∴x=z-2

Now , $\int\left(2 x^{2}+3\right) \cdot \sqrt{x+2} d x$

$=\int\left\{2(z-2)^{2}+3\right\} \sqrt{z} d 2$

$=\int\left\{2\left(z^{2}+4-4z\right)+3\right\} \sqrt{2} d z$

$=\int\left(2 z^{2}+8-8 z+3\right) \sqrt{z} d z$

$=\int\left(2z^{2}+11-8z\right) \sqrt{z} d 2$

$=2 \int2^{\frac{5}{2}} d z+11 \int z^{\frac{1}{2}} d z-8\int z^{\frac{1}{2}} d z$

$=2 \cdot \frac{2}{7} \cdot z^{\frac{7}{2}}+11 \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}-8 \cdot \frac{2}{5} \cdot z^{\frac{3}{2}}+c$

$=2z^{\frac{3}{2}}\left[\frac{2}{7} z^{2}+\frac{11}{3}-\frac{8}{5} z\right]+c$

$=2(x+2)^{\frac{3}{2}}\left[\frac{2}{7}(x+8)^{2}-\frac{8}{5}(x+2)+\frac{11}{3}\right]+c$

Question 13
$\int \frac{x}{\sqrt{x+5}} d x$
Sol :
Let x+5=z then $\frac{d z}{d x}=1$ ∴dx=dz

Again , x+5=z  ∴x=z-5

Now , $\int \frac{x}{\sqrt{x+5}} d x$

$=\int \frac{z-5}{\sqrt{z}} d z$

$=\int \frac{z}{\sqrt{z}} d z-5 \int \frac{1}{\sqrt{z}} d z$

$=\int \sqrt{z} d z-5 \int z^{-\frac{1}{2}} d z$

$=\frac{2}{3} \cdot z^{\frac{3}{2}}-5 \times 2 \cdot z^{-\frac{1}{2}}+c$

$=z^{\frac{1}{2}}\left[\frac{2}{3} z-10\right]+c$

$=\sqrt{x+5} \cdot\left(\frac{2}{3}(x+5)-10\right)+c$

$=\sqrt{x+5}.\left(\frac{2 x+10-30}{3}\right)+c$

$=\sqrt{x+5} \cdot \frac{(2 x-20)}{3}+c$

$=\frac{2}{3}(x-10) \cdot \sqrt{x+5}+c$


Question 14
$\int \frac{x}{\sqrt{1-2 x}} d x$
Sol :
Let z=1-2x then $\frac{d z}{d x}=-2 \therefore d x=-\frac{d z}{2}$

Again , z=1-2x $\therefore x=\frac{1-z}{2}$

Now , $\int \frac{x}{\sqrt{1-2 x}} \cdot d x$ $=\int \frac{1-2}{2 \cdot \sqrt{2}} \cdot\left(-\frac{d z}{2}\right)$

$=\frac{-1}{4} \int \frac{(1-z)}{\sqrt{z}} d 2$

$=\frac{-1}{4} \int \frac{1}{\sqrt{z}} d z+\frac{1}{4}\int\frac{z}{\sqrt{z}} d z$

$=-\frac{1}{4} \int z^{-\frac{1}{2}} d z+\frac{1}{4} \int \sqrt{z} d z$

$=-\frac{1}{4} \cdot 2 \cdot z^{\frac{1}{2}}+\frac{1}{4} \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}+c$

$=-\frac{1}{2} z^{\frac{1}{2}}+\frac{1}{6} z^{\frac{3}{2}}+c$

$=2^{\frac{1}{2}}\left[-\frac{1}{2}+\frac{1}{6} z\right]+c$

$=z^{\frac{1}{2}}\left(\frac{-3+2}{6}\right)$

$=\sqrt{1-2 x} \cdot\left(\frac{-3+1-2 x}{6}\right)+c$

$=\sqrt{1-2 x}\left(\frac{-2-2 x}{6}\right)+c$

$=\frac{-2}{6}(1+x) \sqrt{1-2 x}+c$

$=-\frac{1}{3}(1+x) \sqrt{1-2 x}+c$


Question 15
$\int x \cdot \sqrt{x+2} d x$
Sol :
Let z=x+2 then $\frac{d z}{d x}=1$ ∴dz=dx

Again z=x+2  ∴x=z-2

Now , $\int x \cdot \sqrt{x+2} d x$

$=\int(z-2) \cdot \sqrt{z} d z$

$=\int z^{\frac{3}{2}} d z-2 \int z^{\frac{1}{2}} d z$

$=\frac{2}{5} \cdot z^{\frac{5}{2}}-2 \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}+c$

$=2 z^{\frac{3}{2}}\left[\frac{z}{5}-\frac{2}{3}\right]+c$

$=2z^{\frac{3}{2}}\left[\frac{3z-10}{15}\right]+c$

$=2 \cdot(x+2)^{\frac{3}{2}}\left[\frac{3(x+2)-10}{15}\right]+c$

$=\frac{2}{15}(x+2)^{\frac{3}{2}}\lceil 3 x-4]+c$

Question 16
$\int \frac{2 x+1}{\sqrt{3 x+2}} d x$
Sol :
Let z=3x+2 then $\frac{d z}{d x}=3$ $\therefore d x=\frac{d z}{3}$

Again , z=3x+2  $\therefore \quad x=\frac{z-2}{3}$

Now , $\int \frac{2 x+1}{\sqrt{3 x+2}} d x$

$=\int \frac{2\left(\frac{z-2}{3}\right)+1}{\sqrt{z}} \cdot \frac{d z}{3}$

$=\displaystyle\int \dfrac{2(z-2)+3}{3 \sqrt{z}} \dfrac{dz}{3}$

$=\frac{1}{9} \int \frac{(2z-4+3)}{\sqrt{z}} d z$

$=\frac{1}{9} \int \frac{(2z-1)}{\sqrt{z}} d z$

$=\frac{1}{9} \cdot 2 \int \frac{z}{\sqrt{z}} d z-\frac{1}{9} \int \frac{1}{\sqrt{z}} d z$

$=\frac{2}{9} \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}-\frac{1}{9} \cdot 2 \cdot z^{\frac{1}{2}}+c$

$=2z^{\frac{1}{2}}\left[\frac{2}{27} \cdot z-\frac{1}{9}\right]+c$

$=2 z^{\frac{1}{2}}\left[\frac{2 z-3}{27}\right]+c$

$=\frac{2}{27}(3 x+2)^{\frac{1}{2}} \cdot[2(3 x+8)-3]+c$

$=\frac{2}{27} \cdot(3 x+2)^{\frac{1}{2}}[6 x+4-3]+c$

$=\frac{2}{27} \cdot(6 x+1) \cdot \sqrt{3 x+2}+c$

Question 17
$\int \frac{3 x+5}{\sqrt{7 x+9}} d x$
Sol :
Let z=7x+9 then $\frac{d z}{d x}=7$ $\therefore d x=\frac{d z}{7}$

Again , z=7x+9 $\therefore x=\frac{z-9}{7}$

Now , $\int \frac{3 x+5}{\sqrt{7 x+9}} d x$

$=\int \frac{3\frac{(z-9)}{7}}{\sqrt{z}}+5 \cdot \frac{d z}{7}$

$=\frac{1}{49} \int \frac{3z+8}{\sqrt{2}} d z$

$=\frac{1}{49} \cdot 3 \int \frac{z}{\sqrt{z}} d z+\frac{8}{49} \int \frac{1}{\sqrt{z}} d z$

$=\frac{3}{49} \int \sqrt{z} d 2+\frac{8}{49} \int z^{-\frac{1}{2}} d z$

$=\frac{3}{49} \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}+\frac{8}{49} \cdot 2 \cdot z^{\frac{1}{2}}+c$

$=\frac{2z^{\frac{1}{2}}}{49}[z+8]+c$

$=\frac{2}{49} \cdot \sqrt{7 x+9} \cdot(7 x+9+8)+c$

$=\frac{2}{49} \cdot \sqrt{7 x+9} \cdot(7 x+17)+c$

Question 18
$\int \frac{x^{2}}{\sqrt{x-1}} d x$
Sol :
Let z=x-1 then $\frac{d 2}{d x}=1 \therefore d x=d 2$

Again , z=x-1 $\therefore x=z+1$

Now , $\int \frac{x^{2}}{\sqrt{x-1}} d x$

$=\int \frac{(z+1)^{2}}{\sqrt{Z}} d z$

$=\int \frac{z^{2}+1+2 z}{\sqrt{z}} d z$

$=\int \frac{z^{2}}{\sqrt{z}} d z+\int \frac{1}{\sqrt{z}} d z+2 \int \frac{z}{\sqrt{z}} d z$

$\int z^{\frac{3}{2}} d z+\int z^{-\frac{1}{2}} d z+2 \cdot \int z^{\frac{1}{2}} d z$

$=\frac{2}{5} \cdot z^{\frac{5}{2}}+2 \cdot z^{\frac{1}{2}}+2 \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}+c$

$=2z^{\frac{1}{2}}\left[\frac{z^{2}}{5}+1+\frac{2}{3} z\right]+c$

$=2 \sqrt{z}\left[\frac{3z^{2}+15+10z}{15}\right]+c$

$=\frac{2}{15} \sqrt{x-1}\left[3(x-1)^{2}+15+10(x-1)\right]+c$

$=\frac{2}{15} \cdot \sqrt{x-1}\left[3\left(x^{2}+1-2 x\right)+15+10 x-10\right]+c$

$=\frac{2}{15} \cdot \sqrt{x-1}\left[3 x^{2}+3-6 x+15+10 x-10\right]+c$

$=\frac{2}{15} \cdot \sqrt{x-1} \cdot\left(3 x^{2}+4 x+8\right)+c$

Question 19
$\int \frac{x^{2}}{\sqrt{1-x}} \cdot d x$
Sol :
Let z=1-x then $\frac{d z}{d x}=-1 \quad \therefore d x=-d z$

Again z=1-x $\therefore x=z+1$

Now , $\int \frac{x^{2}}{\sqrt{1-x}} d x$

$=-\int \frac{(z+1)^{2}}{\sqrt{z}} d z$

$=\int \frac{(-z-1)^{2}}{\sqrt{z}} d z$

$=\int \frac{z^{2}+1+2 z}{\sqrt{z}} d z$

$=\int \frac{z^{z}}{\sqrt{z}} d z+\int \frac{1}{\sqrt{z}} d z+2 \int \frac{z}{\sqrt{z}} d z$

$=\int z^{\frac{3}{2}} d z+\int z^{-\frac{1}{2}} d z+2 \int z^{\frac{1}{2}} d z$

$=\frac{2}{5} \cdot 2^{\frac{5}{2}}+2 \cdot z^{\frac{1}{2}}+2 \cdot \frac{2}{3} z^{\frac{3}{2}}+c$

$=2z^{\frac{1}{2}}\left[\frac{z^{2}}{5}+1+\frac{2}{3} z\right]+c$

$=2 \sqrt{z}\left[\frac{3z^{2}+15+10z}{15}\right]+c$

$=\frac{2}{15} \sqrt{z}\left[3(x-1)^{2}+15+10(x-1)\right]+c$

$=\frac{2}{15} \sqrt{2}\left[3\left(x^{2}+1-2 x\right)+15+10 x-10\right]+c$

$=\frac{2}{15} \sqrt{2}\left[3 x^{2}+3-6 x+15+10 x-10\right]+c$

$=\frac{2}{15} \cdot \sqrt{1-x}\left[3 x^{2}+8+4 x\right]+c$

Question 20
$\int \frac{x}{\sqrt{x+4}} d x \quad x>0$
Sol :
Let z=x+4 then $\frac{d z}{d x}=1 \quad \therefore d z=d x$

Again z=x+4 $\therefore x=z-4$

Now , $\int \frac{x}{\sqrt{x+4}} d x$

$=\int \frac{z-4}{\sqrt{z}} \cdot d z$

$=\int \frac{z}{\sqrt{z}} d z-4 \int \frac{1}{\sqrt{z}} d z$

$=\int \sqrt{z} d z-4 \int z^{-\frac{1}{2}} d z$

$=\int z^{\frac{1}{2}} d z-4 \int z^{-\frac{1}{2}} d z$

$=\frac{2}{3} \cdot 2^{\frac{3}{2}}-4 \cdot 2 \cdot z^{\frac{1}{2}}+c$

$=2z^{\frac{1}{2}}\left[\frac{z}{3}-4\right]+c$

$=2 \cdot \sqrt{x+1}\left[\frac{x+1}{3}-4\right]+c$

$=2 \sqrt{x+4}\left(\frac{x+4-12}{3}\right)+c$

$=\frac{2}{3} \cdot \sqrt{x+4} \cdot(x-8)+c$

Question 21
$\int \frac{x^{2}}{\sqrt{3 x+4}} d x$
Sol :
Let z=3x+4 then $\frac{d z}{d x}=3 \therefore d x=\frac{d z}{3}$

Again , z=3x+4 $\therefore x=\frac{z-4}{2}$

Now , $\int \frac{x^{2}}{\sqrt{3 x+1}} d x$

$=\int \frac{\left(\frac{z-4}{3}\right)^{2}}{\sqrt{z}} \frac{d z}{3}$

$=\int \frac{(z-4)^{2}}{9 \cdot \sqrt{z}} \cdot \frac{d z}{3}$

$=\frac{1}{27} \int \frac{z^{2}+16-82}{\sqrt{z}} d 2$

$=\frac{1}{27} \int \frac{z^{2}}{\sqrt{z}} d z+\frac{1}{27} \cdot 16 \int \frac{1}{\sqrt{z}} d z \frac{-8}{27} \int \frac{z d}{\sqrt{z}}$

$=\frac{1}{27} \int z^{\frac{3}{2}} d z+\frac{16}{27} \int z^{-\frac{1}{2}} d z-\frac{8}{27} \int z^{\frac{1}{2}} d z$

$=\frac{1}{27} \cdot \frac{2}{5} \cdot z^{\frac{5}{2}}+\frac{16}{27} \cdot 2z^{\frac{1}{2}}-\frac{8\times2}{27 \times 3} z^{\frac{3}{2}}+c$

$=\frac{2}{27} z^{\frac{1}{8}}\left[\frac{z^{2}}{5}+16-\frac{8}{3} z\right]+c$

$=\frac{2}{27} \sqrt{3 x+4}\left[\frac{(3 x+4)^{2}}{5}-\frac{8}{3}(3 x+4)+16\right]+c$


Question 22
$\int \frac{x^{2}}{\sqrt{x+1}} d x$
Sol :
 Let z=x+1 then $\frac{d z}{d x}=1 \quad \therefore d x=d 2$

Again z=x+1 ∴x=z-1

Now , $\int \frac{x^{2}}{\sqrt{x+1}} d x$

$=\int \frac{(z-1)^{2}}{\sqrt{z}} d z$

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

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

$=\int z^{\frac{3}{2}} d z+\int z^{-\frac{1}{2}} d z-2 \int z^{\frac{1}{2}} d z$

$=\frac{2}{5} \cdot z^{\frac{5}{2}}+2 \cdot z^{\frac{1}{2}}-2 \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}+c$

$=2 z^{\frac{1}{2}}\left[\frac{z^{2}}{5}+1-\frac{2}{3} z\right]+c$

$=2 \sqrt{x+1}\left[\frac{(x+1)^{2}}{5}+1-\frac{2(x+1)}{3}\right]+c$


Question 23
$\int \frac{2 x-1}{(x-1)^{2}} d x$
Sol :
Let z=x-1 then $\frac{d z}{d x}=1 \quad \therefore d z=d x$

Again z=x-1 $\therefore \quad x=z+1$

Now , $\int \frac{2 x+1}{(x-1)^{2}} d x$

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

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

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

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

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

$=2 \cdot \log |z|+\frac{z^{-2+1}}{-2+1}+c$

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

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


Question 24
$\int \frac{x^{2}+3 x+2}{(x-2)(x+1)} d x$
Sol :
$=\int \frac{x^{2}+x+2 x+2}{(x-2)(x+1)} d x$

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

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

$=\int \frac{x+2}{x-2} d x$

$=\int \frac{(x-2)+4}{(x-2)} d x$

$=\int \frac{x+2}{x-2} d x+4 \int \frac{1}{x-2} d x$

$=\int d x+4 \int \frac{1}{x-2} d x$

=x+4log|x+2|+c


Question 25
$\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x$
Sol :
Let z=x+1 then $\frac{d z}{d x}=1 \quad \therefore \quad d z=d x$

Again z=x+1 $\cdots x=z-1$

Now , $\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x$

$=\int \frac{(z-1)^{2}+3(z-1)+1}{2^{z}} d z$

$=\int \frac{z^{2}+1-2z+3z-3+1}{z^{2}} d z$

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

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

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

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

$=z+\log |z|+z^{-1}$

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

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

$=x+\log |x+1|+\frac{1}{x+1}+c$


Question 26
$\int \frac{d x}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$
Sol :
Let $x=2^{6} \therefore z=x^{\frac{1}{6}} \therefore \quad d x=6z^{5} d z$

Now , $\int \frac{d x}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$

$=\int \frac{6 z^{5} d z}{\left(z^{6}\right)^{1/2} t+\left(z^{6}\right)^{1 / 3}}$

$=\int \frac{6z^{5} d z}{z^{3}+z^{2}}$

$=6 \int \frac{z^{5}}{z^{3}+z^{2}} d z$

$=6 \int \frac{z^{5}d z}{z^{2}(z+1)}$

$=6 \int \frac{z^{3}}{z+1} d z$

$=6 \int \frac{z^{3}-1+1}{z+1} d z$

$=6 \int \frac{(z+1)\left(z^{2}+1-2\right)}{(z+1)} d z-6 \int \frac{1}{z+1} d z$

[a³+b³=(a+b)(a²+b²-ab)]

$=6 \int z^{2} d z+6 \int d z-6 \int z d z-6 \log |z+1|+c$

$=6 \cdot \frac{z^{3}}{3}+6z-\frac{6z^{2}}{2}-6 \log |z+1|+c$

$=2z^{3}+6z-3z^{2}-6 \log |z+1|+c$

$=2 \cdot\left(x^{\frac{1}{6}}\right)^{3}+6 \cdot x^{\frac{1}{6}}-3\left(x^{\frac{1}{6}}\right)^{2}-6 \log \left|1+x^{\frac{1}{6}}\right|+c$

$=2 x^{\frac{1}{2}}+6 x^{\frac{1}{6}}-3 x^{\frac{1}{3}}-6 \log \left|1+x^{\frac{1}{6}}\right|+c$

$=2 \sqrt{x}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log \left|1+x^{\frac{1}{8}}\right|+c$