KC Sinha Solution Class 12 Chapter 19 Indefinite Integrals (अनिश्चित समाकल) Exercise 19.12

Exercise 19.12

Question 1

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

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

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

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

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


(ii) $\int \frac{dx}{\sqrt{9+25x^2}}$
Sol :
Given: $I=\int \frac{dx}{\sqrt{9-25x^2}}=\int \frac{dx}{\sqrt{(3)^2-(6x)^2}}$

$\left[\therefore \int \frac{dx}{\sqrt{a^2-x^2}=\sin ^{-1}\frac{x}{a}}\right]$

$I=\frac{1}{5}\sin^{-1} \frac{5x}{3}+C$

Question 2

(i) $\int \frac{dx}{\sqrt{x^2+3x+4}}$
Sol :
Given: $I=\int \frac{dx}{\sqrt{x^2+3x+4}}$

$=\int \dfrac{dx}{\sqrt{x^2+2x\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+4}}$

$=\int \dfrac{dx}{\sqrt{x^2+2x\frac{3}{2}+\frac{9}{4}+\frac{7}{4}}}$

$=\int \dfrac{dx}{\sqrt{\left(x+\frac{3}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}}$

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

$I=\log \left[x+\frac{3}{2}+\sqrt{\left(x+\frac{3}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right]$

$I=\log \left[x+\frac{3}{2}+\sqrt{x^2+3x+4}\right]+C$


(ii) $\int \frac{dx}{\sqrt{2ax-x^2}}$
Sol :
Given :
$I=\int \frac{dx}{\sqrt{2ax-x^2}}=\int \frac{dx}{\sqrt{2ax-x^2-a^2+a^2}}$

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

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

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

$I=\sin ^{-1}\frac{(x-a)}{a}+C$

Question 3

$\int \sqrt{3x^2+4x+1}dx$
Sol :
Given :
$I=\int \sqrt{3x^2+4x+1}dx$

$=\int \sqrt{3x^2+\frac{3}{3}\times 4x+\frac{3}{3}}dx$

$I=\int \sqrt{3\left(x^2+\frac{4x}{3}+\frac{1}{3}\right)}$

$I=\sqrt{3}\int \sqrt{x^2+\frac{4x}{3}+\frac{1}{3}} dx$

$I=\sqrt{3}\int \sqrt{x^2+2.x.\frac{2}{3}+\frac{4}{3}-\frac{4}{3}+\frac{1}{3}}dx$

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

$\left[7\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log |x+\sqrt{x^2-a^2}|\right]$

$I=\sqrt{3}\left[\dfrac{\left(x+\frac{2}{3}\right)}{2}\sqrt{\left(x+\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2}-\frac{1}{9\times 2}\log \left(x+\frac{2}{3}\right)+\sqrt{\left(x+\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2})\right]$

$I=\sqrt{3}\left[\frac{3x+2}{6}\sqrt{\left(x+\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2}-\frac{1}{18}\log \left(x+\frac{2}{3}+\sqrt{\left(x+\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2}\right)\right]$

$I=\sqrt{3}\left[\frac{3x+2}{6}\sqrt{\frac{3x^2+4x+1}{3}}-\frac{1}{18}\log \left(x+\frac{2}{3}\sqrt{\frac{3x^2+4x+1}{3}}\right)\right]+C$

$I=\sqrt{3}\left[\frac{3x+2}{6}\frac{\sqrt{3x^2+4x+1}}{\sqrt{3}}-\frac{1}{18}\log \left(x+\frac{2}{3}+\sqrt{\frac{3x^2+4x+1}{3}}\right)\right]+C$

$I=\frac{3x+2}{6}\sqrt{3x^2+4x+1}-\frac{\sqrt{3}}{18}\log \left(x+\frac{2}{3}+\sqrt{\frac{3x^2+4x+1}{3}}\right)+C$

Question 4

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

$I=\int \sqrt{x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+2}dx$

$I=\int \sqrt{\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)-\frac{1}{4}} dx$

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

$I=\dfrac{\left(x-\frac{3}{2}\right)}{2}\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{8}\right)^2}-\dfrac{\left(\frac{1}{2}\right)^2}{2}\log \left(x-\frac{3}{2}+\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right)+C$

$I=\frac{2x.3}{2\times 2}\sqrt{x^2-3x+2}-\frac{1}{8}\log \left(x-\frac{3}{2}+\sqrt{x^2-3x+2}\right)+C$

$I=\frac{2x.3}{4}\sqrt{x^4-3x+2}-\frac{1}{8}\log \left|\left(x-\frac{3}{2}+\sqrt{x^2-3x+2}\right)\right|+C$

Question 5

$\int \frac{dx}{\sqrt{10-8x-2x^2}}$
Sol :
Given :
$I=\int \frac{dx}{10-8x-2x^2}$

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

$=\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{5-4x-x^2}}$

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

$=\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{9-(x^2+4x+4)}}$

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

$\left[\because \int \frac{dx}{\sqrt{a^2-x^2}}=\sin ^{-1}\frac{x}{a}\right]$

$I=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{x+2}{3}\right)+C$

Question 6

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

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

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

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

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

$=\frac{1}{\sqrt{2}}\int \frac{dx}{\frac{9}{4}-\left(x+\frac{1}{2}\right)^2}$

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

$=\frac{1}{\sqrt{2}}\sin ^{-1}\dfrac{\left(x+\frac{1}{2}\right)}{\frac{3}{2}}+C$

$=\frac{1}{\sqrt{2}}\sin^{-1}\frac{2x+1}{2\times \frac{3}{2}}+C$

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

Question 7

$\int \frac{dx}{\sqrt{16-2x+2x^2}}$
Sol :
Given : $I=\int \frac{dx}{\sqrt{16-2x-2x^2}}$

$=\int \frac{dx}{\sqrt{2(8-x-x^2)}}$

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

$=\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{8(x^2+2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4})}}$

$=\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{8-(x^2+2.x.\frac{1}{2}+\frac{1}{4})+\frac{1}{4}}}$

$=\frac{1}{\sqrt{2}}\int \dfrac{dx}{\sqrt{\frac{33}{4}-\left(x+\frac{1}{2}\right)^2}}$

$=\frac{1}{\sqrt{2}}\int \dfrac{dx}{\sqrt{\left(\frac{\sqrt{33}}{2}\right)^2-\left(x+\frac{1}{2}\right)^2}}$

$=\frac{1}{\sqrt{2}}\sin ^{-1}\dfrac{\left(x+\frac{1}{2}\right)}{\frac{\sqrt{33}}{2}}+C$

$=\frac{1}{\sqrt{2}}\sin ^{-1}\dfrac{(2x+1)}{2\times \frac{\sqrt{33}}{2}}+C$

$I=\frac{1}{\sqrt{2}}\sin ^{-1}\frac{(2x+1)}{\sqrt{33}}+C$

Question 8

$\int \frac{dx}{\sqrt{x^2+2x+2}}$
Sol :
Given :
$I=\int \frac{dx}{\sqrt{x^2+2x+2}}$

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

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

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

$I=\log \left|(x+1)+\sqrt{(x+1)^2+1}\right|+C$

$I=\log \left|(x+1)+\sqrt{x^2+2x+2}\right|+C$

Question 9

$\int \frac{dx}{\sqrt{7-6x-x^2}}$
Sol :
Given : $I=\int \frac{dx}{\sqrt{7-6x-x^2}}$

$=\int \frac{dx}{\sqrt{7-(x^2+6x)}}$

$=\int \frac{dx}{\sqrt{7-(x^2+2.x.3+9-9)}}$

$=\int \frac{dx}{\sqrt{7-(x^2+2.x.3+9)-9}}$

$=\int \frac{dx}{\sqrt{16-(x^2+2.x.3+9)}}$

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

$I=\sin ^{-1}\frac{(x+3)}{4}+C$

Question 10

$\int \frac{dx}{\sqrt{5x^2-2x}}$
Sol :
Given : I$=\int \frac{dx}{\sqrt{5x^2-2x}}$

$=\int \frac{dx}{\sqrt{5(x^2-2.x.\frac{1}{5}+\frac{1}{25}-\frac{1}{25})}}$

$=\frac{1}{\sqrt{5}}\int \frac{dx}{\sqrt{(x^2-2.x.\frac{1}{5}+\frac{1}{25})-\frac{1}{25}}}$

$=\frac{1}{\sqrt{5}}\int \frac{dx}{\sqrt{\left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}}$

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

$I=\frac{1}{\sqrt{5}}\log \left|\left(x-\frac{1}{5}\right)+\sqrt{x^2+\frac{1}{25}-2.x.\frac{1}{5}-\frac{1}{25}}\right|+C$

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

Question 11

$\int \frac{dx}{\sqrt{8+3x+x^2}}$
Sol :
Given : $I=\int \frac{dx}{\sqrt{8+3x+x^2}}$

$I=\int \frac{dx}{\sqrt{8-\left(x^2-2x.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)}}$

$I=\int \frac{dx}{\sqrt{8-\left(x^2-2x.x.\frac{3}{2}+\frac{9}{4}\right)+\frac{9}{4}}}$

$I=\int \frac{dx}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^2}}$

$I=\int \dfrac{dx}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}}$

$I=\sin ^{-1} \dfrac{\left(x-\frac{3}{2}\right)}{\frac{\sqrt{41}}{2}}+C$

$I=\sin^{-1} \dfrac{(2x-3)}{2\times \frac{\sqrt{41}}{2}}+C$

$I=\sin ^{-1}\dfrac{(2x-3)}{\sqrt{41}}+C$

Question 12

$\int \frac{dx}{\sqrt{2x-x^2}}$
Sol :
Given :
$I=\int \frac{dx}{\sqrt{2x-x^2}}$

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

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

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

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

=sin^-1(x-1)+C

Question 13

(i) $\int \frac{dx}{\sqrt{5-4x+x^2}}$
Sol :
Given :
$I=\int \frac{dx}{\sqrt{5-4x+x^2}}$

$=\int \frac{dx}{\sqrt{x^2-2.x.2+4-4+5}}$

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

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

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

$=\log \left|(x-2)+\sqrt{x^2-4x+5}+C\right|$


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

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

$=\int \frac{dx}{\sqrt{x^2-4x+5}}$

$I=\log \left|(x-2)+\sqrt{x^2-4x+5}\right|+C$

Question 14

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

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

$=\int \frac{dx}{\sqrt{x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+2}}$

$=\int \frac{dx}{\sqrt{\left(x^2-2x.\frac{3}{2}+\frac{9}{4}\right)-\frac{1}{4}}}$

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

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

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


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

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

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

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

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

$=\int \frac{dx}{\sqrt{1-\left(x^2-2x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+3\right)}}$

$=\int \frac{dx}{\sqrt{1-\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)+\frac{9}{4}-3}}$

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

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

$=\sin ^{-1} \dfrac{\left(x-\frac{3}{2}\right)}{\frac{1}{2}}+C$

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

=sin^-1(2x-3)+C

Question 15

$\int \frac{dx}{\sqrt{(x-a)(x-b)}}$
Sol :
Given :
$I=\int \frac{dx}{\sqrt{(x-a)(x-b)}}$

$=\int \frac{dx}{\sqrt{x^2-bx-ax+ab}}$

$=\int \frac{dx}{\sqrt{x^2-x(a+b)+ab}}$

$=\int \dfrac{dx}{\sqrt{x^2-2.x.\frac{(a+b)}{2}-\left(\frac{a+b}{2}\right)^2+ab}}$

$=\int \dfrac{dx}{\sqrt{\left(x^2-2.x.\frac{(a+b)}{2}+\left(\frac{a+b}{2}\right)^2\right)-\left(\frac{a+b}{2}\right)^2+ab}}$

$=\int \dfrac{dx}{\sqrt{\left(x-\frac{(a+b)}{2}\right)^2-\frac{(a^2+b^2+2ab)}{4}+ab}}$

$=\int \dfrac{dx}{\sqrt{\left(x-\frac{(a+b)}{2}\right)^2-\left(\frac{a^2+b^2+2ab}{4}\right)^2+ab}}$

$=\int \dfrac{dx}{\left(x-\frac{(a+b)}{2}\right)^2-\left(\frac{(a-b)}{2}\right)^2}$

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

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

Question 16

(i) $\int \frac{dx}{x^2-16}$
Sol :
Given : $I=\int \frac{dx}{x^2-16}$

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

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


(ii) $\int \frac{dx}{x(x^5+3)}$
Sol :
Given : $I=\int \frac{dx}{x(x^5+3)}$

Let x⁵=z

$x=z^{\frac{1}{5}}$

then 5x⁴dx=dz

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

Now ,

$I=\int \frac{dx}{x(x^5+3)}$

$=\int \dfrac{1}{5} \times \frac{dz}{z^{4/5}z^{\frac{1}{5}}(z+3)}$

$=\frac{1}{5}\int \dfrac{dz}{z(z+3)}=\frac{1}{5}\int \frac{dz}{z^2+3z}$

$=\frac{1}{5}\int \dfrac{dz}{\left(z^2+2.z.\frac{3}{2}+\frac{9}{4}\right)-\frac{9}{4}}$

$=\frac{1}{5}\int \dfrac{dz}{\left(z+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2}$

$=\frac{1}{5}\times \dfrac{1}{2\times \frac{3}{2}}\log \left|\dfrac{z+\frac{3}{2}-\frac{3}{2}}{z+\frac{3}{2}+\frac{3}{2}}\right|+C$

$=\frac{1}{15}\log \left|\frac{z}{z+3}\right|+C$

$=\frac{1}{15}\log \left|\frac{x^5}{x^5+3}\right|+C$

Question 17

$\int \frac{dx}{3x^2+13x-10}$
Sol :
Given : $I=\int \frac{dx}{3x^2+13x-10}$

$=\int \dfrac{dx}{3\left(x^2+\frac{13x}{3}-\frac{10}{3}\right)}$

$=\int \frac{1}{3}\int \dfrac{dx}{\left(x^2+2.x.\frac{13}{6}+\frac{169}{36}-\frac{10}{3}\right)}$

$=\frac{1}{3} \int \dfrac{dx}{\left(x+\frac{13}{6}\right)^2-\frac{289}{36}}$

$=\frac{1}{3}\int \frac{dx}{\left(x+\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2}$

$=\frac{1}{3}\times \dfrac{1}{2\times \frac{17}{6}}\log \left|\dfrac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+C$

$=\frac{1}{17}\log \left|\dfrac{x-\frac{4}{6}}{x+5}\right|+C$

$=\frac{1}{17}\log \left|\frac{3x-2}{3(x+5)}\right|+C$

$=\frac{1}{17}$[log(3x-2)-log{3(x+5)}]+C

$=\frac{1}{17}$[log(3x-2)-{log3+log(x+5)}]+C

$=\frac{1}{17}$[log(3x-2)-log3-log(x+5)]+C

$=\frac{1}{17}\log (3x-2)-\frac{1}{17}\log 3-\frac{1}{17}\log (x+5)+C$

$=\frac{1}{17}\log (3x-2)-\frac{1}{17}\log (x+6)+C$

$=\frac{1}{17}$[log(3x-2)-log(x+5)]+C

$I=\frac{1}{17}\log \left|\frac{3x-2}{x+5}\right|+C$

Question 18

$\int \frac{dx}{x^2-6x+13}$
Sol :
Given : $I=\int \frac{dx}{x^2-6x+13}$

$=\int \frac{dx}{x^2-2.x.3+9-9+13}$

$=\int \frac{dx}{(x^2-2.x.3+9)+4}$

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

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

Question 19

(i) $\int \sqrt{4-x^2}$ dx
Sol :
Given : $I=\int \sqrt{4-x^2}$

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

$=\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2}+C$

$I=\frac{x\sqrt{4-x^2}}{2}+\frac{2\sin^{-1}x}{2}+C$


(ii) $\int \sqrt{1-4x^2}$
Sol :
Given : $I=\int \sqrt{1-4x^2}dx$

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

$=\frac{2x}{2}\sqrt{1-4x^2}+\frac{1}{2}\sin^{-1}\frac{(2x)}{1}+C$

$=x\sqrt{1-4x^2}+\frac{1}{2}\sin^{-1}(-2x)+C$

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

Question 20

(i) $\int \sqrt{3-2x-x^2}$
Sol :
Given : $I=\int \sqrt{3-2x-x^2}$

$=\int \sqrt{3-(x^2+2x)}$

$=\int \sqrt{3-(x^2+2.x.3+1-1)}$

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

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

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

$=\frac{(x+1)\sqrt{4-(x+1)^2}}{2}+\frac{4}{2}\sin^{-1}\frac{(x+1)}{2}+C$

$I=\frac{(x+1)}{2}\sqrt{3-2x-x^2}+2\sin^{-1}\left(\frac{x+1}{2}\right)+C$

(ii) $\int \sqrt{1-4x-x^2} dx$
Sol :
Given : $I=\int \sqrt{1-4x-x^2} dx$

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

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

$=\int \sqrt{1-(x^2+2x.2+4)+4}$

$=\sqrt{5-(x^2+2.x.2+4)}$

$=\sqrt{(\sqrt{5})^2-(x+2)^2} dx$

$=\frac{(x+2)}{2}\sqrt{5-(x+2)^2}+\frac{5}{2}\sin^{-1} \left(\frac{x+2}{\sqrt{5}}\right)+C$

$=\frac{(x+2)}{2}\sqrt{1-4x-x^2}+\frac{5}{2}\sin^{-1} \left(\frac{x+2}{\sqrt{5}}\right)+C$

Question 21

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

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

$=\int \sqrt{1-\left(x^2-2x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)}dx$

$=\int \sqrt{1-\left(x^2-2x.\frac{3}{2}+\frac{9}{4}\right)+\frac{9}{4}}dx$

$=\int \sqrt{\frac{13}{4}-\left(x-\frac{3}{2}\right)^2}dx$

$=\int \sqrt{\left(\frac{\sqrt{13}}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}dx$

$=\dfrac{\left(x-\frac{3}{2}\right)}{2}\sqrt{\left(\frac{\sqrt{13}}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}+\frac{13}{4\times 2}\sin^{-1}\dfrac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}}+C$

$=\frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\dfrac{(2x-3)}{\frac{\sqrt{13}}{2}\times 2}+C$

$I=\frac{(2x-3)}{4}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1} \frac{(2x-3)}{\sqrt{13}}+C$

Question 22

$\int \sqrt{x^2+4x-5}dx$
Sol :
Given : $I=\int \sqrt{x^2+4x-5}dx$

$=\int \sqrt{x^2+2.x.2+4-4-5}dx$

$=\int \sqrt{(x^2+2.x.2+4)-9}dx$

$=\int \sqrt{(x+2)^2-(3)^2}$

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

$=\frac{x+2}{2}\sqrt{x^2+4x-5}-\frac{9}{5}\log \left|x+2+\sqrt{x^2+4x-5}\right|+C$

Question 23

$\int \sqrt{x^2+4x+1}dx$
Sol :
Given : $I=\int \sqrt{x^2+4x+1}dx$

$=\int \sqrt{x^2+2x.2+4-4+1}dx$

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

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

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

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

Question 24

(i) $\int \sqrt{x^2+3x}dx$
Sol :
Given : $I=\int \sqrt{x^2+3x}dx$

$=\int \sqrt{\left(x^2+2.x\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)}$

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

$=\dfrac{\left(x+\frac{3}{2}\right)}{2}\sqrt{\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2}-\frac{9}{4\times 2} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2}\right|+C$

$I=\frac{2x+3}{4}\sqrt{x^2+3x}-\frac{9}{8}\log \left|x+\frac{3}{2}+\sqrt{x^2+3x}\right|+C$


(ii) $\int \sqrt{x^2-8x+7}dx$
Sol :
Given : $I=\int \sqrt{x^2-8x+7}dx$

$=\int \sqrt{x^2-2x.4+16-16+7}dx$

$=\int \sqrt{(x^2-2x.4+16)-9}dx$

$=\int \sqrt{(x-4)^2-(3)^2}dx$

$=\frac{(x-4)}{2}\sqrt{(x-4)^2-(3)^2}-\frac{9}{2}\log \left|x-4+\sqrt{(x-4)^2-(3)^2}\right|+C$

$I=\frac{x-4}{2}\sqrt{x^2-8x+7}-\frac{9}{2}\log \left|x-4+\sqrt{x^2-8x+7}\right|+C$

Question 25

(i) $\int \sqrt{1+x^2}dx$
Sol :
Given : $I=\int \sqrt{1+x^2}dx$

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

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

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


(ii) $\int \sqrt{1+\frac{x^2}{9}}dx$
Sol :
Given : $I=\int \sqrt{1+\frac{x^2}{9}}dx$

$=\int \sqrt{\frac{9+x^2}{9}}dx$

$=\int \frac{\sqrt{9+x^2}}{3}dx$

$=\frac{1}{3}\int \sqrt{x^2+9}dx$

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

$=\frac{1}{3}\left[\frac{x}{2}\sqrt{x^2+9}+\frac{9}{2}\log \left|x+\sqrt{x^2+9}\right|\right]+C$

$I=\frac{x}{6}\sqrt{x^2+9}+\frac{3}{2}\log \left|x+\sqrt{x^2+9}\right|+C$

Question 26

$\int \sqrt{x^2+4x+6}dx$
Sol :
Given : $I=\int \sqrt{x^2+4x+6}dx$

$=\int \sqrt{x^2+2.x.2+4-4+6}dx$

$=\int \sqrt{(x^2+2.x.2+4)+2}dx$

$=\int \sqrt{(x+2)^2+(\sqrt{2})^2}dx$

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

$I=\frac{x+2}{2}\sqrt{x^2+4x+6}+\log \left|x+2+\sqrt{x^2+4x+6}\right|+C$

Question 27

$\int \sqrt{x^2+2x+5}dx$
Sol :
Given : $I=\int \sqrt{x^2+2x+5}dx$

$=\int \sqrt{(x^2+2.x.1+1)-1+5}dx$

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

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

$=\frac{x+1}{2}\sqrt{x^2+2x+5}+2\log \left|x+1+\sqrt{x^2+2x+5}\right|+C$

Question 28

$\int \frac{x+3}{x^2-2x+5}dx$
Sol :
Given : $I=\int \frac{x+3}{x^2-2x+5}dx$

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

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

Equating the coefficient of similar power of x
2A=1 and 3=-2A+B

$A=\frac{1}{2}$ and B=3+2A$=3+2\times \frac{1}{2}$=3+1=4


Now , $I=\int \frac{x+3}{x^2-2x-5}dx$

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

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

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

$=\frac{1}{z}\log |z|+4\int \frac{dx}{(x-1)^2-(\sqrt{6})^2}$

$=\frac{1}{2}\log \left|x^2-2x-5\right|+4\left[\frac{1}{2\sqrt{6}}\log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|\right]+C$

$I=\frac{1}{2}\log \left|x^2-2x-5\right|+\frac{2}{\sqrt{6}}\log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+C$

Question 29

$\int \frac{5x-2}{1+2x+3x^2}dx$
Sol :
Given : $I=\int \frac{5x-2}{1+2x+3x^2}dx$

Let 5x-2=A$\frac{d}{dx}$(3x²+2x+1)+B

5x-2=A(6x+2)+B=6Ax+2A+B

Equating the coefficient of similar power of x
6A=5 and 2A+B=-2

$A=\frac{5}{6}$ and B=-2-2A=-2-2×$\frac{5}{6}$

$A=\frac{5}{6}$ and $B=\frac{-11}{3}$


Now , $I=\int \frac{5x-2}{3x^2+2x+1}dx$

$=\int \dfrac{\frac{5}{6}(6x+2)-\frac{11}{3}}{3x^2+2x+1}dx$

$=\frac{5}{6}\int \frac{6x+2}{3x^2+2x+1}dx-\frac{11}{3}\int \frac{dx}{3x^2+2x+1}$

$=\frac{5}{6}\int \frac{dz}{z}-\frac{11}{3} \int \frac{dx}{3\left(x^2+2.x.\frac{1}{3}+\frac{1}{3}\right)}$

$=\frac{5}{6}\log |z|-\frac{11}{3\times 3}\int \frac{dx}{\left(x^2+2.x.\frac{1}{3}+\frac{1}{9}-\frac{1}{9}+\frac{1}{3}\right)}$

$=\frac{5}{6}\log \left|3x^2+2x+1\right|-\frac{11}{9}\int \dfrac{dx}{\left(x+\frac{1}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)}$

$=\frac{5}{6}\log \left|3x^2+2x+1\right|-\frac{11}{9}\times \dfrac{1}{\frac{\sqrt{12}}{3}}\tan^{-1}\dfrac{\left(x+\frac{1}{3}\right)}{\frac{\sqrt{2}}{3}}+C$

$=\frac{5}{6}\log \left|3x^2+2x+1\right|-\frac{11}{3\sqrt{2}}\tan^{-1} \dfrac{\left(3x+1\right)}{3\times \frac{\sqrt{2}}{3}}+C$

$I=\frac{5}{6} \log \left|3x^2+2x+1\right|-\frac{11}{3\sqrt{2}}\tan ^{-1}\frac{(3x+1)}{\sqrt{2}}+C$

Question 30

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

Let x+2=A$\frac{d}{dx}$(2x²+6x+5)

x+2=A(4x+6)+B=4Ax+6A+B

Equating the coefficient of similar power of x
4A=1 and 6A+B=2

$A=\frac{1}{4}$ and

B=2-6A$=2-6\times \frac{1}{4}$

$=2-\frac{3}{2}=\frac{1}{2}$


Now ,  $I=\int \frac{x+2}{2x^2+6x+5}dx$

$=\int \dfrac{\frac{1}{4}(4x+6)+\frac{1}{1}}{2x^2+6x+5}dx$

$=\frac{1}{4}\int \frac{4x+6}{2x^2+6x+5}dx+\frac{1}{2}\int \frac{dx}{2x^2+6x+5}$

$=\frac{1}{4}\int \frac{dz}{z}+\frac{1}{2}\int \dfrac{dx}{2(x^2+3x+\frac{5}{2})}$

$=\frac{1}{4}\int \frac{dz}{z}+\frac{1}{2\times 2}\int \dfrac{dx}{\left(x^2+2x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+\frac{5}{2}\right)}$

$=\frac{1}{4}\log~z+\frac{1}{4}\int \dfrac{dx}{\left(x+\frac{3}{2}\right)^2+\frac{1}{4}}$

$=\frac{1}{4} \log~z+\frac{1}{4}\int \dfrac{dx}{\left(x+\frac{3}{8}\right)^2+\left(\frac{1}{2}\right)^2}$

$=\frac{1}{4}\log \left|2x^2+6x+5\right|+\frac{1}{4}\dfrac{1}{\frac{1}{2}}\tan^{-1} \dfrac{\left(x+\frac{3}{2}\right)}{\frac{1}{2}}+C$

$=\frac{1}{4}\log \left|2x^2+6x+5\right|+\frac{2}{4}\tan^{-1}\dfrac{2x+3}{2\times \frac{1}{2}}+C$

$I=\frac{1}{4}\log \left|2x^2+6x+5\right|+\frac{1}{2}\tan^{-1} (2x+3)+C$

Question 31

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

Let 3x+1=A$\frac{d}{dx}$(2x²-2x+3)+B
3x+1=A(4x-2)+B=4Ax-2A+B

Equating the coefficient of similar power of x
4A=3 and

-2A+B=1
B=1+2A

∴$A=\frac{3}{4}$ and $B=1+2\times \frac{3}{4}=1+\frac{3}{2}=\frac{5}{2}$

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

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

$=\int \dfrac{\frac{3}{4}(4x-2)+\frac{5}{2}}{2x^2-2x+3}dx$

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

$=\frac{3}{4}\int \frac{dz}{z}+\frac{5}{2}\frac{dx}{2(x^2-x+\frac{3}{2})}$

$=\frac{3}{4}\int \frac{1}{z}dz+\frac{5}{2\times 2}\int \frac{dx}{\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+\frac{3}{2}\right)}$

$=\frac{3}{4}\log \left|2x^2-2x+3\right|+\frac{5}{4}\int \dfrac{dx}{\left(x-\frac{1}{2}\right)^2+\frac{5}{4}}$

$=\frac{3}{4}\log \left|2x^2-2x+3\right|+\frac{5}{4}\int \dfrac{dx}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{5}}{2}\right)^2}$

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

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

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

Question 32

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

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

Equating the coefficient of similar power of x
2A-1 and B=-1
∴$A=\frac{1}{2}$

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

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

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

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

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

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

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

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

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

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


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

Let x+2=A$\frac{d}{dx}$(x²-1)+B=A(2x)+B
x+2=2A(x)+B
∴2A=1 and B=2

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


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

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

$=\int \dfrac{\frac{1}{2}\times 2x+2}{\sqrt{x^2-1}}dx$

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

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

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

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

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

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

Question 33

$\int \frac{x+2}{\sqrt{4x-x^2}}dx$
Sol :
Given : $I=\int \frac{x+2}{\sqrt{4x-x^2}}dx$

Let x+2A.$\frac{d}{dx}(4x-x^2)$+B=A(4-2x)+B

x+2=4A-2Ax+B

Equating the coefficient of similar power of x
we get -2A=1 and 4A+B=2

$A=\frac{1}{2}$ B=2-4A=$2-4\times \left(-\frac{1}{2}\right)$

B=2+2=4


Now , $I=\int \frac{x+2}{\sqrt{4x-x^2}}dx$

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

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

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

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

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

$=-\frac{1}{2}\dfrac{z^{\frac{1}{2}}}{\frac{1}{2}}+4\sin^{-1}\frac{(x-2)}{2}+C$

$-\sqrt{z}+4\sin^{-1}\frac{(x-2)}{2}+C$

$I=-\sqrt{4x-x^2}+4\sin^{-1}(x-2)+C$

Question 34

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

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

x+1=2A-2Ax+B

Equating the coefficient of similar power of x
We get 1=-2A and B+2A=1

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


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

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

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

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

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

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

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

$=-\frac{1}{2}\times \dfrac{z^{\frac{1}{2}}}{\frac{1}{2}}+2\sin^{-1}\frac{(x-1)}{1}+C$

=-√z+2sin^-1(x-1)+C
=-√2x-x²+2sin^-1(x-1)+C

Question 35

$\int \frac{2x+1}{\sqrt{7+6x-3x^2}}$
Sol :
Given : $I=\int \frac{2x+1}{\sqrt{7+6x-3x^2}}$

Let 2x+1=A$\frac{d}{dx}(7+6x-3x^2)+B$ =A(6-6x)+B

2x+1=6A-6Ax+B

Equating the coefficient of similar power of x
We get -6A=2 and 6A+B=1

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

B=1-6A$=1-6\times \left(-\frac{1}{3}\right)$
=1+2=3


Now , $I=\int \frac{2x+1}{\sqrt{7+6x-3x^2}}$

$=\int \frac{A(6-6x)+B}{\sqrt{7+6x-3x^2}}$

$=\int \frac{A(6-6x)}{\sqrt{7+6x-3x^2}}dx+\int \frac{B}{\sqrt{7+6x-3x^2}}dx$

$=A\int \frac{(6-6x)dx}{\sqrt{7+6x-3x^2}}+B\int \frac{dx}{\sqrt{7+6x-3x^2}}$

$=-\frac{1}{3}\int \frac{dz}{\sqrt{z}}+3\int \frac{dx}{\sqrt{7+3-3+6x-3x^2}}$

$=-\frac{1}{3}2\sqrt{z}+3\int \frac{10}{\sqrt{10-(3x^2-6x+3)}}$

$=-\frac{2}{3}\sqrt{z}+3\int \frac{dx}{\sqrt{(\sqrt{10})^2-(\sqrt{3}x-\sqrt{3})^2}}$

$=-\frac{2}{3}\sqrt{7+6x-3x^2}+3\frac{\sin^{-1}(\sqrt{3}x-\sqrt{3})}{\sqrt{3}\times \sqrt{10}}+C$

$I=-\frac{2}{3}\sqrt{7+6x-3x^2}+\sqrt{3}\frac{\sin^{-1}\sqrt{3}(x-1)}{\sqrt{10}}+C$

Question 36

$\int \frac{4x+1}{\sqrt{2x^2+x-3}}dx$
Sol :
Given : $I=\int \frac{4x+1}{\sqrt{2x^2+x-3}}dx$

Let z=2x²+x-3 then dz=(4x+1)dx

Now , $I=\int \frac{4x+1}{\sqrt{2x^2+x-3}}dx=\int \frac{dz}{\sqrt{z}}$

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

$=\dfrac{z^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C$

$=2\sqrt{z}+C$

$=2\sqrt{2x^2+x-3}+C$

Question 37

$\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx$
Sol :
Given : $I=\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx$

Let 5x+3=A$\frac{d}{dx}(x^2+4x+10)$ =A(2x+4)+B

5x+3=2Ax+4A+B

Equating the coefficient of similar power of x

We get 2A=5 and 4A+B=3

$A=\frac{5}{2}$ and

B=3-4A$=3-4\times \frac{5}{2}$

B=3-10=-7

Now , $I=\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx$

$=\int \frac{A(2x+4)+B}{\sqrt{x^2+4x+10}}$

$=A\int \frac{(2x+4)}{\sqrt{x^2+4x+10}}dx+B\int \frac{dx}{\sqrt{x^2+4x+10}}$

$I=\frac{5}{2}\int \frac{(2x+4)}{\sqrt{x^2+4x+10}}dx-7\int \frac{dx}{\sqrt{x^2+4x+4+6}}$

$=\frac{5}{2}\int \frac{dz}{\sqrt{z}}-7\int \frac{dx}{\sqrt{x^2+2.x.2+4+6}}$

$I=\frac{5}{2}.2\sqrt{z}-7\int \frac{dx}{\sqrt{(x+2)^2+(\sqrt{6})^2}}$

$=5\sqrt{z}-7\log \left|x+2+\sqrt{(x+2)^2+(\sqrt{6})^2}\right|+C$

$I=5\sqrt{x^2+4x+10}-7\log \left|(x+2)+\sqrt{x^2+4x+10}\right|+C$

Question 38

$\int \frac{x+3}{\sqrt{5-4x-x^2}}$
Sol :
Given : $I=\int \frac{x+3}{\sqrt{5-4x-x^2}}$

Let x+3=A$\frac{d}{dx}(5-4x-x^2)+B$=A(-4-2x)+B

x+3=-4A-2Ax+B=-2Ax+B-4A

Equating the coefficient of similar power of x
We get -2A=1 and B-4A=3

∴$A=-\frac{1}{2}$

B=3+4A$=3+4\times -\frac{1}{2}$
=3-2=1


Now , $I=\int \frac{x+3}{\sqrt{5-4x-x^2}}$

$=\int \frac{A(-4-2x)+B}{\sqrt{5-4x-x^2}}dx$

$=\int \frac{A(-4-2x)}{\sqrt{5-4x-x^2}}dx+\int \frac{B}{\sqrt{5-4x-x^2}}dx$

$=A\int \frac{(-4-2x)}{\sqrt{5-4x-x^2}}dx+B\int \frac{dx}{\sqrt{5-4x-x^2}}$

$I=-\frac{1}{2}\int \frac{dz}{\sqrt{z}}+1\int \frac{dx}{\sqrt{5+4-4-4x-x^2}}$

$=-\frac{1}{2}\times 2\sqrt{z}+\int \frac{dx}{\sqrt{9-(x^2+4x+4)}}$

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

$=-\sqrt{5-4x-x^2}+\sin^{-1}\frac{(x+2)}{3}+C$

Question 39

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

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

x+2=2Ax-2A+B

Equating the coefficient of similar power of x
We get 1=2A and B-2A=2
∴$A=\frac{1}{2}$

B=2+2A$=2+2\times \frac{1}{2}=3$


Now , $I=\int \frac{x+2}{\sqrt{x^2-2x+3}}dx$

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

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

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

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

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

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

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

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

Question 40

$\int \frac{6x+7}{(x-5)(x-4)}dx$
Sol :
Given : $I=\int \frac{6x+7}{(x-5)(x-4)}dx$

$= \int \frac{6x+7}{\sqrt{(x^2-4x-5x+20)}}$

$=\int \frac{6x+7}{\sqrt{x^2-9x+20}}dx$

Let 6x+7=A$\frac{d}{dx}(x^2-9x+20)+B$=A(2x-9)+B

6x+7=2Ax-9A+B

Equating the coefficient of similar power of x
We get , 2A=6 and B-9A=7
A=3  B=9A+7=27+7=34


Now , $I=\int \frac{6x+7}{\sqrt{x^2-9x+20}}$

$=\int \frac{A(2x-9)+B}{\sqrt{x^2-9x+20}}dx$

$=A\int \frac{dz}{\sqrt{z}}+B\int \frac{dx}{\sqrt{\left(x^2-2.x.\frac{9}{2}+\frac{81}{4}\right)-\frac{81}{4}+20}}$

$I=3\times 2\sqrt{2}+34\int \frac{dx}{\sqrt{\left(x-\frac{9}{2}\right)^2-\frac{1}{4}}}$

$=6\sqrt{z}+34\int \frac{dx}{\sqrt{\left(x-\frac{9}{8}\right)^2-\left(\frac{1}{2}\right)^2}}$

$=6\sqrt{x^2-9x+20}+34\log \left|\left(x-\frac{9}{8}\right)+\sqrt{\left(x-\frac{9}{8}\right)^2-\left(\frac{1}{2}\right)^2}\right|+C$

$=6\sqrt{x^2-9x+20}+34\log \left|x-\frac{9}{8}+\sqrt{x^2-9x+20}\right|+C$

Question 41

(i) $\int (x+2)\sqrt{x^2+1}dx$
Sol :
Given : $I=\int (x+2)\sqrt{x^2+1}dx$

$=\int x\sqrt{x^2+1}dx+2\int \sqrt{x^2+1}dx$

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

$=\frac{1}{2}\int \sqrt{z}dz+2\left[\frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\log \left|x+\sqrt{x^2+1}\right|\right]+C$

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


(ii) $\int (x+1) \sqrt{2x^2+3}dx$
Sol :
Given : $I=\int (x+1) \sqrt{2x^2+3}dx$

$=\int x.\sqrt{2x^2+3}dx+\int \sqrt{2x^2+3}dx$

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

Let z=2x²+3 then dz=4xdx

Now , $I=\frac{1}{4}\int 4x\sqrt{2x^2+3}dx+\sqrt{2}\int \sqrt{x^2+\frac{3}{2}}dx$

$=\frac{1}{4}\int \sqrt{2x^2+3}4xdx+\sqrt{2} \int \sqrt{(x)^2+\left(\sqrt{\frac{3}{2}}\right)^2}dx$

$=\frac{1}{4}\int \sqrt{z}dz+\sqrt{2}\left[\frac{x}{2}\sqrt{(x)^2+\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^2}+\frac{3}{2\times 2}\log \left|x+\sqrt{x^2+\frac{3}{2}}\right|\right]+C$

$=\frac{1}{4}\times \frac{2}{3}z^{\frac{3}{2}} +\sqrt{2}\left[\frac{x}{2}\sqrt{x^2+\frac{\sqrt{3}}{\sqrt{2}}}+\frac{3}{4}\log \left|x+\sqrt{x^2+\frac{3}{2}}\right|\right]+C$

$=\frac{1}{6}(2x^2+3)^{3/2}+\sqrt{2}\left[\frac{x}{2}\sqrt{\frac{2x^2+3}{2}}+\frac{3}{4}\log \left|x+\sqrt{x^2+\frac{3}{2}}\right|\right]$

$=\frac{1}{6}(2x^2+3)^{3/2}+\sqrt{2}\left[\frac{x}{2}\sqrt{\frac{2x^2+3}{\sqrt{2}}}+\frac{3}{4}\log \left|x+\sqrt{x^2+\frac{3}{2}}\right|\right]$

$=\frac{1}{6}(2x^2+3)^{3/2}+\frac{x}{2}\sqrt{2x^2+3}+\frac{3\sqrt{2}}{4}\log\left|x+\sqrt{x^2+\frac{3}{2}}\right| +C$

Question 42

$\int x.\sqrt{x+x^2}dx$
Sol :
Given : $I=\int x.\sqrt{x+x^2}dx$

$=\frac{1}{2}\int 2x.\sqrt{x+x^2}dx$

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

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

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

$=\frac{1}{2}\int \sqrt{z}dz-\frac{1}{2}\int \sqrt{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}]\right)^2}dx$

$=\frac{1}{2}\times \frac{2}{3}z^{\frac{3}{2}}-\frac{1}{2}\left[\dfrac{\left(x+\frac{1}{2}\right)}{2}\sqrt{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}-\frac{1}{4\times 2}\log \left|x+\frac{1}{2}+\sqrt{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right|\right]+C$

$=\frac{1}{3}z^{\frac{3}{2}}-\frac{1}{2}\left[\frac{2x+1}{2\times 2}\sqrt{x^2+x}-\frac{1}{8}\log \left|x+\frac{1}{2}+\sqrt{x^2+x}\right|\right]+C$

$=\frac{1}{3}z^{3/8}-\frac{(2x+1)}{8}\sqrt{x^2+x}+\frac{1}{16}\log \left|x+\frac{1}{2}+\sqrt{x^2+x}\right|+C$

$I=\frac{1}{3}(x^2+x)^{3/2}-\frac{1}{8}(2x+1)\sqrt{x^2+x}+\frac{1}{16}\log \left|x+\frac{1}{2}+\sqrt{x^2+x}\right|+C$

Question 43

(i) $\int (x+3)\sqrt{3-4x-x^2}dx$
Sol :
Given : $I=\int (x+3)\sqrt{3-4x-x^2}dx$

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

$=\frac{1}{2}\int (2x+6)\sqrt{3-4x-x^2}dx$

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

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

Let z=3-4x-x² then dz=-4-2x=-(2x+4)dx

∴-dz=(2x+4)dx

Now $I=\frac{1}{2}\int \sqrt{z}(-dz)+\int \sqrt{3-(x^2+4x)}dx$

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

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

$=-\frac{1}{3}z^{3/2}+\int \sqrt{7-(x^2+4x+4)}dx$

$=-\frac{1}{3}z^{3/2}+\int \sqrt{(\sqrt{7})^2-(x^2)^2}dx$

$-\frac{1}{3}(3-4x-x^2)^{3/2}+\frac{(x+2)}{2}\sqrt{(\sqrt{7})^2-(x+2)^2}+\frac{(\sqrt{7})^2}{2}\sin ^{-1}\frac{(x+2)}{7}+C$

$I=-\frac{1}{3}(3-4x-x^2)^{3/2}+\frac{(x+2)}{2}\sqrt{3-4x-x^2}+\frac{7}{2}\sin^{-1}\frac{(x+2)}{\sqrt{7}}+C$


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

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

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

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

Let z=1+x-x² then dz=(1-2x)dx=-(2x-1)dx
-dz=(2x-1)dx

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

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

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

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

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

$=-\frac{1}{3}z^{3/2}+\frac{1}{4}\int \sqrt{5-(4x^2-4x+1)}dx$

$=-\frac{1}{3}z^{3/2}+\frac{1}{4}\int \sqrt{(\sqrt{5})^2-(2x-1)^2}dx$

$=-\frac{1}{3}z^{3/2}+\frac{1}{4}\left[\frac{(2x-1)}{2}\sqrt{(\sqrt{5})^2-(2x-1)^2}+\frac{5}{2\times 2}\sin^{-1}\frac{2x-1}{\sqrt{5}}\right]$

$=-\frac{1}{3}(1+x-x^2)^{3/2}+\frac{(2x-1)}{8}\sqrt{1+x-x^2}+\frac{5}{16}\sin^{-1}\frac{(2x-1)}{\sqrt{5}}+C$

Question 44

$\int \frac{3x^2}{1+x^6}dx$
Sol :
Given : $I=\int \frac{3x^2}{1+x^6}dx$

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

Let x³=z then 3x²dx=dz

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

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

I=tan^-1(z)+C
=tan^-1(x³)+C

Question 45

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

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

Let x³=z then 3x²dx=dz⇒x²dx$=\frac{dz}{3}$

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

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

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

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

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

Question 46

$\int \frac{3x}{1+2x^4}dx$
Sol :
Given : $I=\int \frac{3x}{1+2x^4}dx$

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

Let $z=\sqrt{2}x^2$ then $dz=\sqrt{2}2xdx$⇒$\frac{dz}{2\sqrt{2}=xdx}$


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

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

$=\frac{1\times 3}{2\sqrt{2}}\int \frac{1}{1+z^2}dz$

$=\frac{3}{2\sqrt{2}}\times \frac{1}{1}\tan^{-1}\frac{z}{1}+C$

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

Question 47

$\int \frac{x^2}{1-9x^6}dx$
Sol :
Given : $I=\int \frac{x^2}{1-9x^6}dx$

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

Let $z=3x^3$ then $dz=3.3x^2dx=9x^2dx$⇒$\frac{dz}{9}=x^2dx$


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

$=\int \frac{dz}{9(1-z^2)}$

$=\frac{1}{9}\int \frac{dz}{1-z^2}$

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

$=\frac{1}{9}\times \frac{1}{2\times 1} \log \left|\frac{1+z}{1-z}\right|+C$

$I=\frac{1}{18}\log \left|\frac{1+3x^3}{1-3x^3}\right|+C$

Question 48

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

$=\int \frac{x^2}{\sqrt{(x^3)^2+(a^3)^2}}dx$

Let $z=x^3$ then $dz=3x^2dx$⇒$\frac{dz}{3}=x^2dx$

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

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

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

$=\frac{1}{3}\log \left|z+\sqrt{z^2+(a^3)^2}\right|+C$ $\left[\therefore \int \frac{dx}{\sqrt{x^2+a^2}=\log \left|x+\sqrt{x^2+a^2}\right|}\right]$

$I=\frac{1}{3}\log \left|x^3+\sqrt{x^6+a^6}\right|+C$

Question 49

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

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

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

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

$=\frac{1}{2}\int \frac{dz}{z}-\frac{1}{2}\int \frac{dx}{(1+\left(x^2+2.x.\frac{1}{2}+\frac{1}{4})-\frac{1}{4}\right)}$

$=\frac{1}{2}\log |z|-\frac{1}{2}\int \dfrac{dx}{\frac{3}{2}+\left(x+\frac{1}{2}\right)^2}$

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

$=-\frac{1}{\sqrt{3}}\tan^{-1}\dfrac{2x+1}{2\times \frac{\sqrt{3}}{2}}+C+\frac{1}{2}\log |x^2+x+1|+C$

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

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

Question 50

$\int \frac{dx}{x(x^3+8)}$
Sol :
Given : $I=\int \frac{dx}{x(x^3+8)}$

Let $z=x^3$ then $dz=3x^2dx$⇒$\frac{dx}{3x^2}=dx$
∴$x=z^{\frac{1}{3}}$ and ∴$dz=\frac{dz}{3.z^{2/3}}$
∴$x=z^{\frac{2}{3}}$

Now , $I=\frac{dx}{x(x^3+8)}$

$=\int \frac{dz}{3.z^{2/3}.z^{1/3}(z+8)}$

$=\frac{1}{3}\int \frac{dz}{z(z+8)}$

$=\frac{1}{3}\int \frac{dz}{z^2+8z}$

$=\frac{1}{3}\int \frac{dz}{z^2+2.z.4+16-16}$

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

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

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

$=\frac{1}{24}\log \left|\frac{z}{z+8}\right|+C$

$=\frac{1}{24}\log \left|\frac{x^3}{x^3+8}\right|+C$

Question 51

$\int \frac{x+2}{\sqrt{x^2+5x+6}}dx$
Sol :
$I=\int \frac{x+2}{\sqrt{x^2+5x+6}}dx$

Let x+2=A$\frac{d}{dx}(x^2+5x+6)+B$=A(2x+5)+B

x+2=2Ax+5A+B

Equating the coefficient of similar power of x
We get 2A=1 and B+5A=2

∴$A=\frac{1}{2}$ B=2-5A$=2-5\times \frac{1}{2}$ $=2-\frac{5}{2}=\frac{4-5}{2}=\frac{-1}{2}$

Now , $I=\int \frac{x+2}{\sqrt{x^2+5x+6}}dx$

$=\int \frac{A(2x+5)+B}{\sqrt{x^2+5x+6}}dx$

$=A\int \frac{(2x+5)}{\sqrt{x^2+5x+6}}dx+B\int \frac{1}{\sqrt{x^2+5x+6}}$

$=\frac{1}{2}\int \frac{dz}{\sqrt{z}}-\frac{1}{2}\int \frac{1}{\sqrt{\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}\right)-\frac{25}{4}+6}}$

$=\frac{1}{2}\times 2\sqrt{z}-\frac{1}{2}\int \dfrac{dx}{\sqrt{\left(x+\frac{5}{2}\right)^2-\frac{1}{4}}}$

$=\sqrt{z}-\frac{1}{2}\int \dfrac{dx}{\sqrt{\left(x+\frac{5}{2}\right)^2-\left(\frac{1}{2}\right)^2}}$

$=\sqrt{x^2+5x+6}-\frac{1}{2}\log \left|\left(x+\frac{5}{2}\right)+\sqrt{\left(x+\frac{5}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right|+C$

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

$I=\sqrt{x^2+5x+6}-\frac{1}{2}\log \left|\left(x+\frac{5}{2}\right)+\sqrt{x^2+5x+6}\right|+C$