# NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.4

Exercise 7.4

Integrate the following functions.

$$\textbf{1.}\space\frac{\textbf{3x}^\textbf{2}}{\textbf{x}^\textbf{6}\textbf{+1}}\\\textbf{Sol.}\space \text{Let}\space \text{I}=\int\frac{3x^2}{x^6+1}dx\\=\int\frac{3x^2}{(x^3)^2+1}dx\\\text{Let \space} x^3=t\\\Rarr\space 3x^2=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{3x^2}\\\therefore\space \text{I}=\int\frac{3x^2}{t^2+1}.\frac{dt}{3x^2}\\=\int\frac{dt}{t^2+1}$$

$$=\frac{1}{1}\text{tan}^{\normalsize-1}\bigg(\frac{t}{1}\bigg)+\text{C}\\=\text{tan}^{\normalsize-1}(x^3)+\text{C}$$

$$\textbf{2.}\space\frac{\textbf{1}}{\sqrt{\textbf{1+ 4x}^\textbf{2}}}\textbf{.}\\\textbf{Sol.}\space\int\frac{dx}{\sqrt{1+4x^2}}=\int\frac{dx}{\sqrt{4\bigg(\frac{1}{2}\bigg)^2 + x^2}}\\=\frac{1}{2}\int\frac{dx}{\sqrt{x^2 + \bigg(\frac{1}{2}\bigg)^2}}\\=\frac{1}{2}\text{log}\begin{vmatrix}x+\sqrt{x^2 + \bigg(\frac{1}{2}\bigg)^2}\end{vmatrix}+\text{C}\\\bigg(\because\space\int\frac{1}{\sqrt{x^2+a^2}}dx=\text{log}|x + \sqrt{x^2 + a^2}|\bigg)$$

$$=\frac{1}{2}\text{log}\begin{vmatrix}x + \frac{\sqrt{4x^2+1}}{2}\end{vmatrix}+\text{C}\\=\frac{1}{2}\text{log}|2x + \sqrt{4x^2+1}|-\frac{1}{2}\text{log 2} + \text{C}\\\bigg[\because\space\text{log}\bigg(\frac{m}{n}\bigg)=\text{log m - log n}\bigg]\\=\frac{1}{2}\text{log}|2x + \sqrt{4x^2+1}|+\text{C}\\\bigg(\because\space\frac{1}{2}\text{log 2 is constant + constant = C}\bigg)$$

$$\textbf{3.}\space\frac{1}{\sqrt{(2-x)^2+1}}\\\textbf{Sol.}\space\int\frac{1}{\sqrt{(2-x)^2+1}}dx\\=\int\frac{1}{\sqrt{(x-2)^2+1^2}}dx\\=\text{log}|(x-2) + \sqrt{(x-2)^2+1}+\text{C}|\\\bigg[\because\space \int\frac{dx}{\sqrt{x^2 + a^2}}=\text{log}|x+\sqrt{x^2+a^2}|\bigg]$$

$$\textbf{4.}\space\frac{\textbf{1}}{\sqrt{\textbf{9-25}}\textbf{x}^\textbf{2}}\\\textbf{Sol.}\space\int\frac{1}{\sqrt{9-25}x^2}dx\\=\int\frac{1}{\sqrt{25\bigg[\bigg(\frac{3}{5}\bigg)^2 - x^2\bigg]}}dx\\=\frac{1}{5}\int\frac{1}{\sqrt{\bigg[\bigg(\frac{3}{5}\bigg)^2-x^2\bigg]}}dx\\=\frac{1}{5}\text{sin}^{\normalsize-1}\bigg(\frac{x}{\frac{3}{5}}\bigg)+\text{C}$$

$$\bigg[\because\space\int\frac{dx}{\sqrt{a^2-x^2}}=\text{sin}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]\\=\frac{1}{2}\text{sin}^{\normalsize-1}\bigg(\frac{5x}{3}\bigg)+\text{C}$$

$$\textbf{5.\space}\frac{\textbf{3x}}{\textbf{1+2x}^\textbf{4}}\\\textbf{Sol.}\space\int\frac{3x}{1+2x^4}dx\\=\frac{3}{2}\int\frac{xdx}{\frac{1}{2} + x^4}\\=\frac{3}{2}\int\frac{x dx}{\frac{1}{2} + (x^2)^2}\\\text{Let\space x}^2=t\\\Rarr\space 2x=\frac{dt}{dx}\\\Rarr dx=\frac{dt}{2x}\\\therefore\space\frac{3}{2}\int\frac{xdx}{\frac{1}{2} + (x^2)^2}=\frac{3}{2}\int\frac{x}{\frac{1}{2} + (t)^2}\frac{dt}{2x}$$

$$=\frac{3}{2}\int\frac{dt}{t^2 + \bigg(\frac{1}{\sqrt{2}}\bigg)^2}\\=\frac{3}{4}×\frac{1}{\frac{1}{\sqrt{2}}}\text{tan}^{\normalsize-1}\bigg(\frac{t}{\frac{1}{\sqrt{2}}}\bigg)+\text{C}\\\bigg[\because\space\int\frac{dx}{a^2 + x^2}=\frac{1}{a}\text{tan}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]\\=\frac{3}{2\sqrt{2}}\text{tan}^{\normalsize-1}(\sqrt{2} x^2)+\text{C}\\(\because\space t=x^2)$$

$$\textbf{6.}\space\frac{\textbf{x}^\textbf{2}}{\textbf{1-x}^\textbf{6}}.\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{x^2}{1-x^6}dx\\=\int\frac{x^2}{1-(x^3)^2}dx\\\text{Let}\space x^3=t\\\Rarr\space 3x^2=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{3x^2}\\\therefore\space \text{I}=\int\frac{x^2}{1-t^2}×\frac{dt}{3x^2}\\=\frac{1}{3}\int\frac{dt}{1-t^2}$$

$$=\frac{1}{3}.\frac{1}{2}\text{log}\begin{vmatrix}\frac{1+t}{1-t}\end{vmatrix}+\text{C}\\\bigg[\because\space\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\text{log}\begin{vmatrix}\frac{\text{a+x}}{\text{a-x}}\end{vmatrix}\bigg]\\=\frac{1}{6}\space\text{log}\begin{vmatrix}\frac{1+x^3}{1-x^3}\end{vmatrix}+\text{C}\\(\because\space t=x^3)$$

$$\textbf{7.\space}\frac{\textbf{x-1}}{\sqrt{\textbf{x}^\textbf{2}\textbf{-1}}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{x-1}{\sqrt{x^2-1}}dx\\=\int\frac{x}{\sqrt{x^2-1}}dx-\int\frac{x}{\sqrt{x^2-1}}dx$$

= I1 – I2

$$\text{Now,}\space\text{I}_1=\int\frac{x}{\sqrt{x^2-1}}dx,\\\text{Let x}^2-1=t\\\Rarr 2x=\frac{dt}{dx}\Rarr\space dx=\frac{dt}{2x}\\\therefore\space\text{I}_1=\int\frac{x}{\sqrt{t}}\frac{dt}{2x}=\frac{1}{2}\int\frac{dt}{\sqrt{t}}\\=\frac{1}{2}\int t^{\frac{1}{2}}dt\\=\frac{1}{2}\bigg[\frac{t^{1/2}}{1/2}\bigg]+\text{C}_1\\=\sqrt{t}+\text{C}_1\\=\sqrt{x^2-1}+\text{C}_1\\(\because\space t= x^2-1)$$

$$\text{Now,}\space\text{I}_2=\int\frac{1}{\sqrt{x^2-1}}dx\\=\text{log}|x+\sqrt{x^2-1}|+\text{C}_2\\\bigg[\because\space\int\frac{dx}{\sqrt{x^2-a^2}}=\text{log}|x + \sqrt{x^2-a^2}|\bigg]$$

On putting the values of I1 and I2 in Eq. (i), we get

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

(where, C = C1 – C2)

$$\textbf{8.}\space\frac{\textbf{x}^\textbf{2}}{\sqrt{\textbf{x}^\textbf{6} \textbf{+ a}^\textbf{6}}}\\\textbf{Sol.}\space\int\frac{\text{x}^\text{2}}{\sqrt{x^6+a^6}}dx\\=\int\frac{x^2}{\sqrt{(x^3)^2 + a^6}}dx\\\text{Let}\space x^2=t\\\Rarr\space 3x^2=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{3x^2}\\\therefore\space\int\frac{x^2}{\sqrt{(x^3)^2 + a^6}}dx\\=\int\frac{x^2}{\sqrt{t^2 + a^2}}\frac{dt}{3x^2}$$

$$=\frac{1}{3}\int\frac{dt}{\sqrt{t^2 + (a^3)^2}}\\=\frac{1}{3}\text{log}|t+\sqrt{t^2 + a^6}|+\text{C}\\\bigg[\because\space\int\frac{dx}{\sqrt{x^2+a^2}}=\text{log}|x+\sqrt{x^2+a^2}|\bigg]\\=\frac{1}{3}\text{log}|x^3 +\sqrt{x^6 + a^6}|+\text{C}\\(\because\space t=x^3)$$

$$\textbf{9.}\space\frac{\textbf{sec}^\textbf{2}\textbf{x}}{\sqrt{\textbf{tan}^\textbf{2}\textbf{x + 4}}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{\text{sec}^2x}{\text{tan}^2x + 4}dx\\\text{Let}\space\text{tan x = t}\\\Rarr\space\text{sec}^2x=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{\text{sec}^2x}\\\therefore\space\int\frac{\text{sec}^2x}{\sqrt{tan^2x +4}}dx\\=\int\frac{\text{sec}^2x}{\sqrt{t^2+4}}\frac{dt}{\text{sec}^2x}\\=\int\frac{dt}{\sqrt{t^2 + 2^2}}$$

$$=\text{log}|t+\sqrt{t^2 + 4}|+\text{C}\\\bigg[\because\space\int\frac{8x}{\sqrt{x^2 + a^2}}=\text{log}|x + \sqrt{x^2 + a^2}|\bigg]\\=\text{log}|\text{tan x} + \sqrt{\text{tan}^2x+4}| +\text{C}$$

( t = tan x)

$$\textbf{10.}\space\frac{\textbf{1}}{\sqrt{\textbf{x}^\textbf{2}\textbf{ + 2x+2}}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{1}{\sqrt{x^2+2x+2}}dx\\=\int\frac{1}{(x^2 + 2x+1)+1}dx\\=\int\frac{1}{\sqrt{(x+1)^2}+1}dx$$

$$\text{Let x + 1 = t}\Rarr\space dx=dt\\\therefore\space\text{I}=\int\frac{1}{\sqrt{t^2+1}}dt=\text{log}|t + \sqrt{t^2+1}|+\text{C}\\\bigg[\because\space\int\frac{dx}{x^2+a^2}=\text{log}|x+\sqrt{x^2+a^2}|\bigg]\\=\text{log}|(x+1) + \sqrt{(x+1)^2+1}+\text{C}|\\\Rarr\space\text{log}|(x+1) +|\sqrt{x^2+2x+2}+\text{C}\\(\because\space t=x+1)$$

$$\textbf{11.}\space\frac{\textbf{1}}{\textbf{ax}^\textbf{2}\textbf{+ 6x + 5}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{1}{9x^2 + 6x+5}dx\\=\frac{1}{9}\int\frac{1}{x^2+\frac{2}{3}x+\frac{5}{9}}dx\\=\frac{1}{9}\int\frac{1}{x^2+2.\frac{1}{3}.x+\bigg(\frac{1}{3}\bigg)^2 + \frac{5}{9}-\bigg(\frac{1}{3}\bigg)^2}dx\\=\frac{1}{9}\int\frac{1}{x^2 + \frac{2}{3}x+\bigg(\frac{1}{3}\bigg)^2 + \frac{5}{9}-\frac{1}{9}}dx\\=\frac{1}{9}\int\frac{1}{\bigg(x +\frac{1}{3}\bigg)^2 +\bigg(\frac{2}{3}\bigg)^2}dx$$

$$=\frac{1}{9}\bigg(\frac{1}{\frac{2}{3}}\bigg)\text{tan}^{\normalsize-1}\space\bigg(\frac{x+\frac{1}{3}}{\frac{2}{3}}\bigg)+\text{C}\\\bigg[\because\space\int\frac{dx}{a^2+x^2}=\frac{1}{a}\text{tan}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]\\=\frac{1}{6}\text{tan}^{\normalsize-1}\bigg(\frac{3x+1}{2}\bigg)+\text{C}$$

$$\textbf{12.}\space\frac{\textbf{1}}{\sqrt{\textbf{7 - 6x - x}^\textbf{2}}}\\\textbf{Sol.}\space\text{Let \qquad I}=\int\frac{1}{\sqrt{7-6x-x^2}}dx\\=\int\frac{1}{\sqrt{7-(x^2+6x)}}dx\\=\int\frac{1}{\sqrt{7-[x^2+2×3x+(3)^2-(3)^2]}}dx\\ =\int\frac{1}{\sqrt{7-[x^2 + 6x + 3^2-9]}}dx\\=\int\frac{1}{\sqrt{7-[(x+3)^2-9]}}dx\\=\int\frac{1}{\sqrt{4^2 - (x+3)^2}}dx$$

$$\text{Let\space x+3=t}\\\Rarr\space dx=dt\\\therefore\space \text{I}=\int\frac{1}{\sqrt{4^2-t^2}}dt\\=\text{sin}^{\normalsize-1}\bigg(\frac{t}{4}\bigg)+\text{C}\\\bigg[\because\space \int\frac{dx}{\sqrt{a^2-x^2}}=\text{sin}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]\\=\text{sin}^{\normalsize-1}\bigg(\frac{x+3}{4}\bigg)+\text{C}\\(\because\space t= x+3)$$

$$\textbf{13.}\space\frac{1}{\sqrt{\textbf{(x-1)(x-2)}}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{1}{\sqrt{ (x-1)(x-2)}}dx\\=\int\frac{1}{\sqrt{x^2-3x+2}}dx\\=\int\frac{1}{\sqrt{x^2-3x + \bigg(\frac{3}{2}\bigg)^2-\bigg(\frac{3}{2}\bigg)^2+2}}dx\\=\int\frac{1}{\sqrt{\bigg(x-\frac{3}{2}\bigg)^2 + \frac{(-9+8)}{4}}}$$

$$=\int\frac{1}{\sqrt{\bigg(x-\frac{3}{2}\bigg)^2-\bigg(\frac{1}{2}\bigg)^2}}\\\text{Let x}=\frac{3}{2}=t\\\Rarr\space dx=dt\\\therefore\space\text{I}=\int\frac{1}{\sqrt{t^2-\bigg(\frac{1}{2}\bigg)^2}}dt\\=\text{log}\begin{vmatrix} t +\sqrt{t^2-\frac{1}{2}}\end{vmatrix}+\text{C}\\\bigg[\because\space\int\frac{dx}{\sqrt{x^2-a^2}}=\text{log}|x + \sqrt{x^2-a^2}|\bigg]$$

$$=\space \text{log}\begin{vmatrix}x-\frac{3}{2} + \sqrt{\bigg(x-\frac{3}{2}\bigg)^2-\frac{1}{4}}\end{vmatrix}+\text{C}\\=\text{log}\begin{vmatrix}\bigg(x-\frac{3}{2}\bigg) + \sqrt{x^2-3x+2}\end{vmatrix}+\text{C}$$

$$\textbf{14.}\space\frac{\textbf{1}}{\sqrt{\textbf{8+3x-x}^\textbf{2}}}\\\textbf{Sol.}\space\text{Let}\space\qquad\text{I}=\int\frac{1}{\sqrt{8+3x-x^2}}dx\\=\int\frac{1}{\sqrt{8-\bigg[x^2-3x + \bigg(\frac{3}{2}\bigg)^2-\bigg(\frac{3}{2}\bigg)^2\bigg]}}dx\\=\int\space\frac{1}{\sqrt{8-\bigg[\bigg(x-\frac{3}{2}\bigg)^2-\frac{9}{4}\bigg]}}dx\\=\int\frac{1}{\sqrt{8+\frac{9}{4}-\bigg(x-\frac{3}{2}\bigg)^2}}dx$$

$$=\int\frac{1}{\sqrt{\bigg(\frac{\sqrt{41}}{2}\bigg)^2 -\bigg(x-\frac{3}{2}\bigg)^2}}dx\\\text{Let x−}\frac{3}{2}=t\\\Rarr\space dx=dt\\\therefore\space \text{I}=\int\frac{1}{\sqrt{\bigg(\frac{\sqrt{41}}{2}\bigg)^2-t^2}}dt\\=\text{sin}^{\normalsize-1}\bigg(\frac{t}{\frac{\sqrt{41}}{2}}\bigg)+\text{C}\\\bigg[\because\space\int\frac{dx}{\sqrt{a^2-x^2}}=\text{sin}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]$$

$$=\text{sin}^{\normalsize-1}\bigg(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\bigg)+\text{C}\\=\text{sin}^{\normalsize-1}\bigg(\frac{2x-3}{\sqrt{41}}\bigg)+\text{C}\\\bigg(\because\space t=x-\frac{3}{2}\bigg)$$

$$\textbf{15.}\space\frac{\textbf{1}}{\sqrt{\textbf{(x-a)(x-b)}}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{1}{(x-a)(x-b)}dx\\=\int\frac{1}{\sqrt{x^2-(a+b)x+ab}}dx\\=\\\int\frac{1}{\sqrt{\bigg[x^2-(a+b)x+\bigg(\frac{a+b}{2}\bigg)^2-\bigg(\frac{a+b}{2}\bigg)^2+ ab\bigg]}}dx$$

$$=\int\frac{1}{\sqrt{\bigg[x-\bigg(\frac{a+b}{2}\bigg)\bigg]^2 +ab -\bigg(\frac{a+b}{2}\bigg)^2}}dx\\=\int\frac{1}{\sqrt{\bigg[x-\bigg(\frac{a+b}{2}\bigg)\bigg]^2+ab \space -\bigg(\frac{a^2+b^2+2ab}{4}\bigg)}}dx\\=\int\frac{1}{\sqrt{\bigg[x-\bigg(\frac{a+b}{2}\bigg)\bigg]^2 + \frac{4ab-a^2-b^2-2ab}{4}}}dx\\=\int\frac{1}{\sqrt{\bigg[x-\bigg(\frac{a+b}{2}\bigg)^2\bigg] + \frac{2ab -a^2-b^2}{4}}}dx$$

$$=\int\frac{1}{\sqrt{\bigg[x-\bigg(\frac{a+b}{2}\bigg)\bigg]^2-\bigg(\frac{a^2+b^2-2ab}{4}\bigg)}}dx\\=\int\frac{1}{\sqrt{\bigg(x-\frac{a+b}{2}\bigg)-\bigg(\frac{a-b}{2}\bigg)^2}}dx\\\text{Let}\space x-\frac{a+b}{2}=t\\\Rarr\space dx=dt\\\therefore\space \text{I}=\int\frac{1}{\sqrt{t^2-\bigg(\frac{a-b}{2}\bigg)^2}}dt$$

$$=\space\text{log}\begin{vmatrix}t+\sqrt{t^2 -\bigg(\frac{a-b}{2}\bigg)^2}\end{vmatrix}+\text{C}\\\bigg[\because\space\int\frac{dx}{\sqrt{x^2-a^2}}=\text{log}|x+\sqrt{x^2-a^2}|\bigg]$$

$$=\text{log}\begin{vmatrix}\because\space\int\frac{dx}{\sqrt{x^2-a^2}}=\text{log}|x + \sqrt{x^2-a^2}|\end{vmatrix}\\=\text{log}\space\begin{vmatrix}\bigg(x-\frac{a+b}{2}\bigg) + \sqrt{\bigg(x-\frac{a+b}{2}\bigg)^2-\bigg(\frac{a-b}{2}\bigg)^2}\end{vmatrix}\\+\text{C}\\\bigg(\because\space t=x-\frac{(a+b)}{2}\bigg)$$

$$=\\\space\text{log}\begin{vmatrix}\bigg(x-\frac{a+b}{2}\bigg) + \sqrt{x^2 + \frac{(a+b)^2}{4}-x(a+b)-\frac{(a-b)^2}{4}}\end{vmatrix}+\text{C}\\=\text{log}\begin{vmatrix}\bigg(x-\frac{a+b}{2}\bigg) + \sqrt{(x-a)(x-b)}\end{vmatrix}+\text{C}$$

$$\textbf{16.\space}\frac{\textbf{4x+1}}{\sqrt{\textbf{2x}^\textbf{2}\textbf{+x-3}}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{4x+1}{2x^2+x-3}dx\\\text{Let}\space 2x^2+x-3=t\\\Rarr 4x+1=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{4x+1}\\\therefore\space \text{I}=\int\frac{4x+1}{\sqrt{t}}×\frac{dt}{4x+1}\\=\int\frac{1}{\sqrt{t}}dt=2\sqrt{t}+\text{C}\\=2\sqrt{2x^2+x-3}+\text{C}\\(\because\space t= 2x^2+x-3)$$

$$\textbf{17.}\space \frac{\textbf{x+2}}{\sqrt{\textbf{x}^\textbf{2}\textbf{-1}}}\\\textbf{Sol.}\space \int\frac{\text{x+2}}{\sqrt{x^2-1}}dx\\=\int\frac{x}{\sqrt{x^2-1}}dx+\int\frac{2}{\sqrt{x^2-1}}dx\\=I_1+I_2\\\text{Now,}\space\text{I}_1=\int\frac{x}{\sqrt{x^2-1}}dx,$$

Let x2 – 1 = t

$$\Rarr\space 2x dx= dt\\\Rarr\space dx=\frac{dt}{2x}\\\therefore\space \text{I}_1=\int\frac{x}{\sqrt{t}}×\frac{dt}{2x}\\=\frac{1}{2}\int\frac{dt}{\sqrt{t}}\\=\frac{1}{2}\int t^{-1/2}dt\\=\frac{1}{2}[2 t^{1/2}]=\sqrt{t}=\sqrt{x^2-1}+\text{C}_1\\(\because\space t = x^2-1)\\\text{Now,}\space\text{I}_2=2\int\frac{1}{\sqrt{x^2-1}}dx\\ = 2 log x|x + \sqrt{x^2-1}|+\text{C}_2$$

$$\bigg[\because\space\int\frac{dx}{\sqrt{x^2-a^2}}=\text{log}|x + \sqrt{x^2-a^2}|\bigg]$$

On putting the values of I1 and I2 in Eq. (i), we get

$$\text{I}=\sqrt{x^2-1}+2\text{log}|x + \sqrt{x^2+1}|+\text{C}\\\text{where}\space\text{C}=\text{C}_1+\text{C}_2$$

$$\textbf{18.}\space\frac{\textbf{5x-2}}{\textbf{1+2x+3x}^2}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{5x-2}{1+2x+3x^2}dx\\\text{Let}\space 5x-2=\text{A}\frac{d}{dx}(1+2x+3x^2)+\text{B}$$

$$\Rarr\space\text{5x-2} = A(2 + 6x)+B\\\Rarr\space 5x-2=6Ax + (2A+B)$$

On equating the coefficient of x and constant on both sides, we get

5 = 6A

$$\Rarr\space A=\frac{5}{6}\space\text{and}\space 2A+B=-2\\\Rarr\space\text{B}=-\frac{11}{3}\\\therefore\space 5x-2=\frac{5}{6}(2+6x)+\bigg(-\frac{11}{3}\bigg)\\\therefore\space\text{I}=\int\frac{\frac{5}{6}(2+6x)-\frac{11}{3}}{1+2x+3x^2}dx\\=\frac{5}{6}\int\frac{2+6x}{1+2x+3x^2}dx\\-\frac{11}{3}\int\frac{1}{1+2x+3x^2}dx$$

$$\text{Let}\space \text{I}_1=\int\frac{2+6x}{1+2x+3x^2}dx\\\text{and}\space\text{I}_2=\int\frac{1}{1+2x+3x^2}dx\\\therefore\space \text{I}=\frac{5}{6}\text{I}_1-\frac{11}{3}\text{I}_2\space\text{...(i)}\\\text{Now,}\space\text{I}_1=\int\frac{2+6x}{1+2x+3x^2}dx$$

Let 1 + 2x + 3x2 = t

$$\Rarr\space (2+6x)dx=dt\\\therefore\space\text{I}_1=\int\frac{dt}{t}=\text{log}|t|+\text{C}_1\\\Rarr\space\text{I}_1=\text{log}|1+2x+3x^2|+\text{C}_1\space\text{...(ii)}\\\text{Also},\space\text{I}_2=\int\frac{1}{1+2x+3x^2}dx\\\text{1+2x+3x}^2\space\text{can be written as 1+3}\bigg(x^2+\frac{2}{3}x\bigg)\\\text{Therefore}\space 1+3\bigg(x^2+\frac{2}{3}x\bigg)\\=1+3\bigg(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}\bigg)\\=1+3\bigg(x+\frac{1}{3}\bigg)^2-\frac{1}{3}$$

$$=\frac{2}{3}+3\bigg(x+\frac{1}{3}\bigg)^2\\=3\bigg[\bigg(x+\frac{1}{3}\bigg)^2+\frac{2}{9}\bigg]\\=3\bigg[\bigg(x+\frac{1}{3}\bigg)^2+\bigg(\frac{\sqrt{2}}{3}\bigg)^2\bigg]\\\therefore\space\text{I}_2=\frac{1}{3}\int\frac{dx}{\bigg[\bigg(x+\frac{1}{3}\bigg)^2 + \bigg(\frac{\sqrt{2}}{3}\bigg)^2\bigg]}\\=\frac{1}{3}\begin{bmatrix}\frac{1}{\frac{\sqrt{2}}{3}}\text{tan}^{\normalsize-1}\bigg(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\bigg)\end{bmatrix}+\text{C}_2\\\bigg[\because\space\int\frac{dx}{a^2+x^2}=\frac{1}{a}\text{tan}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]$$

$$=\frac{1}{3}\bigg[\frac{3}{\sqrt{2}}\text{tan}^{\normalsize-1}\bigg(\frac{3x+1}{\sqrt{2}}\bigg)\bigg]+\text{C}_2\\=\frac{1}{\sqrt{2}}\text{tan}^{\normalsize-1}\bigg(\frac{3x+1}{\sqrt{2}}\bigg)+\text{C}_2\space\text{...(iii)}$$

On substituting the values of I1 and I2 from equations (ii) and (iii) in Eq. (i), we get

$$\text{I}=\frac{5}{6}[\text{log}|1+2x++3x^2|]-\\\frac{11}{3}\bigg[\frac{1}{\sqrt{2}}\text{tan}^{\normalsize-1}\bigg(\frac{3x+1}{\sqrt{2}}\bigg)\bigg]+\text{C}\\\bigg(\because\space \frac{5}{6}\text{C}_1-\frac{11}{3}\text{C}_2=\text{C}\bigg)\\=\frac{5}{6}\text{log}|1+2x+3x^2|-\\\frac{11}{3\sqrt{2}}\text{tan}^{\normalsize-1}\bigg(\frac{3x+1}{\sqrt{2}}\bigg)+\text{C}$$

$$\textbf{19.}\space\frac{\textbf{6x+7}}{\sqrt{\textbf{(x-5)(x-4)}}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{6x+7}{\sqrt{(x-5)(x-4)}}dx\\=\int\frac{6x+7}{\sqrt{x^2-9x+20}}dx\\\text{Let}\space 6x+7=A\frac{d}{dx}(x^2-9x+20)+\text{B}$$

$$\Rarr\space 6x+7=A(2x-9)+\text{B}\\\Rarr\space 6x+7=2Ax + (-9A+B)$$

On equating the coefficients of x and constant term on both sides, we get

$$\text{2A = 6}\\\Rarr\space A=3\space\text{and}-9A+B=7\\\Rarr\space B=34\\\Rarr\space 6x+7=3(2x-9)+34\\\therefore\space \text{I}=\int\frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}dx\\=3\int\frac{2x-9}{\sqrt{x^2-9x+20}}dx+\\34\int\frac{1}{\sqrt{x^2-9x+20}}dx\\\text{Let}\space\text{I}_1=\int\frac{2x-9}{\sqrt{x^2-9x+20}}dx\\\text{and}\space \text{I}_2=\int\frac{1}{\sqrt{x^2-9x+20}}dx$$

$$x^2-9x+20\space\text{can be written as}\\\space x^2-9x+20+\frac{81}{4}-\frac{81}{4}\\\text{Therefore\space,}x^2-9x+20+\frac{81}{4}-\frac{81}{4}\\=\bigg(x-\frac{9}{2}\bigg)^2-\frac{1}{4}\\=\bigg(x-\frac{9}{2}\bigg)^2-\bigg(\frac{1}{2}\bigg)^2\\\therefore\space\text{I}_2=\int\frac{1}{\bigg(x-\frac{9}{2}\bigg)^2-\bigg(\frac{1}{2}\bigg)^2}dx$$

$$=\text{log}\begin{vmatrix}\bigg(x-\frac{9}{2}\bigg) + \sqrt{\bigg( x-\frac{9}{2}\bigg)^2-\frac{1}{4}}\end{vmatrix}\\+\text{C}_2\\\bigg[\because\space\int\frac{dx}{\sqrt{x^2-a^2}}=\text{log}|x+\sqrt{x^2-a^2}|\bigg]\\=\text{log}\begin{vmatrix}\bigg(x-\frac{9}{2}\bigg) + \sqrt{x^2-9x+20}\end{vmatrix}\\+\text{C}_2\text{...(iii)}$$

On substituting the values of I1 and I2 from equations (ii) and (iii) in equation (i), we get

$$\text{I}=3[2\sqrt{x^2-9x+20}]\\+34\space\text{log}\begin{vmatrix}\bigg(x-\frac{9}{2}\bigg) + \sqrt{x^2-9x+20}\end{vmatrix}+\text{C}\\\lbrack\because\space 3 C_1+34C_2=C\rbrack\\=6\sqrt{x^2-9x+20}+\\34\text{log}\begin{vmatrix}\bigg(x-\frac{9}{2}\bigg)+\\\sqrt{x^2-9x+20}\end{vmatrix}+\text{C}$$

$$\textbf{20}\space\frac{\textbf{x+2}}{\sqrt{\textbf{4x-x}^\textbf{2}}}\\\textbf{Sol.}\space \text{Let}\space x+2=A\frac{d}{dx}(4x-x^2)+B$$

$$\Rarr\space x+2=A(4-2x)+B\\\Rarr\space x+2=-2Ax+4A+B$$

On equating the coefficients of x and constant term on both sides, we get

$$-2A=1\Rarr\space\text{A}=-\frac{1}{2}\space\text{and}\space 4A + B =2\\\Rarr\space B=4\\\Rarr\space (x+2)=-\frac{1}{2}(4-2x)+4\\\therefore\space\int\frac{x+2}{\sqrt{4x-x^2}}dx=\\\int\frac{-\frac{1}{2}(4-2x)+4}{\sqrt{4x-x^2}}dx\\=-\frac{1}{2}\int\frac{4-2x}{\sqrt{4x-x^2}}dx+4\int\frac{dx}{\sqrt{4x-x^2}}$$

$$\text{Let}\space\text{I}_1=\int\frac{4-2x}{\sqrt{4x-x^2}}dx\\\text{and}\space\text{I}_2=\int\frac{dx}{\sqrt{4x-x^2}}\\\text{Then}\space\int\frac{x+2}{\sqrt{4x-x^2}}dx=-\frac{1}{2}\text{I}_1 + 4\text{I}_2\space\text{...(i)}\\\text{Now,}\space \text{I}_1=\int\frac{4-2x}{\sqrt{4x-x^2}}dx$$

Let 4x – x2 = t

$$\Rarr\space (4-2x)dx=dt\\\Rarr\space \text{I}_1=\int\frac{dt}{\sqrt{t}}=2\sqrt{t}+\text{C}_1\\=2\sqrt{4x-x^2}+\text{C}_1\space\text{...(ii)}\\\text{and}\space\text{I}_2=\int\frac{dx}{\sqrt{4x-x^2}}$$

[ 4x – x2 = – (x2 – 4x) = – [(x – 2)2 – 4]

= [(2)2 – (x – 2)2]

$$\therefore\space\text{I}_2=\int\frac{1}{\sqrt{(2)^2-(x-2)^2}}dx\\=\text{sin}^{\normalsize-1}\bigg(\frac{x-2}{2}\bigg)+\text{C}_2\space\text{...(iii)}\\\bigg[\because\space\int\frac{dx}{\sqrt{a^2-x^2}}=\text{sin}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]$$

On substituting the values of I1 and I2 from Eqs. (ii) and (iii) in Eq. (i), we get

$$\int\frac{x+2}{\sqrt{4x-x^2}}dx\\=-\frac{1}{2}[2\sqrt{4x-x^2}]+4\text{sin}^{\normalsize-1}\bigg(\frac{x-2}{2}\bigg)+\text{C}\\\bigg(\because\space -\frac{1}{2}\text{C}_1+4\text{C}_2=\text{C}\bigg)\\=\sqrt{4x-x^2}+4\space\text{sin}^{\normalsize-1}\bigg(\frac{x-2}{2}\bigg)+\text{C}$$

$$\textbf{21.}\space\frac{\textbf{x+2}}{\sqrt{\textbf{x}^\textbf{2}\textbf{+2x+3}}}\\\textbf{Sol.}\space\text{Let x+2}=A\frac{d}{dx}(x^2+2x+3)+B$$

$$\Rarr\space x+2=A(2x+2)+B\\\Rarr\space x+2=2Ax+(2A+B)$$

On equating the coefficient of x and constant term on both sides, we get

$$2A=1\\\Rarr\space A=\frac{1}{2}\space\text{and}\space 2A+B=2\\\Rarr\space 2×\frac{1}{2}+B=2\\\Rarr\space B=2-1=1\\\Rarr\space x+2=\frac{1}{2}(2x+2)+1\\\therefore\space\int\frac{x+2}{\sqrt{x^2+2x+3}}dx\\=\int\frac{\frac{1}{2}(2x+2)+1}{\sqrt{x^2+2x+3}}dx\\=\frac{1}{2}\int\frac{2x+2}{\sqrt{x^2+2x+3}}dx+\\\int\frac{dx}{\sqrt{x^2+2x+3}}$$

$$\text{Let\space}I_1=\int\frac{2x+2}{\sqrt{x^2+2x+3}}dx\\\text{and}\space\text{I}_2=\int\frac{dx}{\sqrt{x^2+2x+3}}\\\text{Then}\space\int\frac{x+2}{\sqrt{x^2+2x+3}}dx\\=\frac{1}{2}\text{I}_1+\text{I}_2\space\text{...(i)}\\\text{Now,}\space\text{I}_1=\int\frac{2x+2}{\sqrt{x^2+2x+3}}dx$$

Let x2 + 2x + 3 = t

$$\Rarr\space(2x+2)dx=dt$$

$$\therefore\space \text{I}_1=\int\frac{dt}{\sqrt{t}}=\int t^{-1/2}dt\\=\frac{t^{(-1/2)+1}}{-\frac{1}{2}+1}+\text{C}_1\\=2\sqrt{t}+\text{C}_1\\=2\sqrt{x^2+2x+3}+\text{C}_1$$

$$\text{and}\space\text{I}_2=\int\frac{dx}{\sqrt{x^2+2x+3}}\\=\int\frac{dx}{\sqrt{x^2+2x+(1)^2+3-(1)^2}}\\=\int\frac{dx}{\sqrt{(x+1)^2+(\sqrt{2})^2}}$$

Let  x + 1 = t

$$\Rarr\space dx=dt\\\text{I}_2=\int\frac{dt}{\sqrt{t^2 + (\sqrt{2})^2}}\\=\text{log}|t+\sqrt{t^2+2}|+\text{C}_2\\\bigg[\because\space\int\frac{dx}{\sqrt{x^2+a^2}}=\text{log}|x+\sqrt{x^2+a^2}|\bigg]$$

$$=\text{log}|x + 1 + \sqrt{(x+1)^2+2}|+\text{C}_2\\=\text{log}|x+1+\sqrt{x^2+2x+3}|+\text{C}_2$$

On putting the values of I1 and I2 eq. (i), we get

$$\int\frac{x+2}{\sqrt{x^2+2x+3}}dx\\=\frac{1}{2}[2\sqrt{x^2+2x+3}]+\text{log}|x+1 +\\\sqrt{x^2+2x+3}|+\text{C}\\\bigg[\because\frac{1}{2}\text{C}_1+4\text{C}_2=\text{C}\bigg]\\=\sqrt{x^2+2x+3}+\\\text{log}|x+1+\sqrt{x^2+2x+3}|+\text{C}$$

$$\textbf{22.}\space\frac{\textbf{x+3}}{\textbf{x}^\textbf{2}\textbf{-2x-5}}\\\textbf{Sol.}\space\text{Let}(x+3)\\=A\frac{d}{dx}(x^2-2x-5)+\text{B}$$

$$\Rarr\space(x+3)=A(2x-2)+B\\\Rarr\space x+3=2Ax-2A+B$$

On equating the coefficients of x and constant term on both sides, we get

2A = 1

$$\Rarr\space\text{A}=\frac{1}{2}\space\text{and}-2A+B=3\\\Rarr\space B=4\\\Rarr\space (x+3)=\frac{1}{2}(2x-2)+4\\\therefore\space\int\frac{x+3}{x^2+2x-5}dx\\=\int\frac{\frac{1}{2}(2x-2)+4}{x^2-2x-5}dx\\=\frac{1}{2}\int\frac{(2x-2)}{x^2-2x-5}dx+\\4\int\frac{1}{x^2-2x-5}dx\\\text{Let}\space\text{I}_1=\int\frac{2x-2}{x^2-2x-5}dx$$

$$\text{and}\space\text{I}_2=\int\frac{1}{x^2-2x-5}dx\\\therefore\space\int\frac{x+3}{x^2+2x+5}dx\\=\frac{1}{2}\text{I}_1+4\text{I}_2\\\text{Now,}\space\text{I}_1=\int\frac{2x-2}{x^2-2x-5}dx$$

Let x2 – 2x – 5 = t

$$\Rarr\space (2x-2)dx=dt\\\therefore\space\text{I}_1=\int\frac{dt}{t}=\text{log}|t|+\text{C}_1\\=\text{log}|x^2-2x-5|+\text{C}_1\space\text{...(ii)}\\\text{and}\space\text{I}_2=\int\frac{1}{x^2-2x-5}dx\\=\int\frac{1}{(x^2-2x+1)-6}dx\\=\int\frac{1}{(x-1)^2-(\sqrt{6})^2}dx\\=\frac{1}{2\sqrt{6}}\text{log}\begin{vmatrix}\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\end{vmatrix}\space\text{...(iii)}\\\bigg[\because\space\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\text{log}\begin{vmatrix}\frac{x-a}{x+a}\end{vmatrix}\bigg]$$

On substituting the values I1 and I2 from Eqs. (ii) and (iii) in Eq. (i), we get

$$\int\frac{x+3}{x^2-2x-5}dx\\=\frac{1}{2}\text{log}|x^2-2x-5|+\\4\bigg[\frac{1}{2\sqrt{6}}\text{log}\begin{vmatrix}\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\end{vmatrix}\bigg]+\text{C}\\\bigg[\because\frac{1}{2}\text{C}_1+4\text{C}_2=\text{C}\bigg]\\=\frac{1}{2}\text{log}|x^2-2x-5|+\\\frac{2}{\sqrt{6}}\text{log}\begin{vmatrix}\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\end{vmatrix}+\text{C}$$

$$\textbf{23.}\space\frac{\textbf{5x+3}}{\sqrt{\textbf{x}^\textbf{2}\textbf{+4x+10}}}\\\textbf{Sol.}\space\text{Let 5x+3}\\=A\frac{d}{dx}(x^2+4x+10)+B\\\Rarr\space 5x+3=A(2x+4)+B\\\Rarr\space 5x+3=2Ax+4A+B$$

On equating the coefficients of x and constant term on both sides, we get

$$2A=5\Rarr\space A=\frac{5}{2}\\\text{and}\space 4A+B=3\\\Rarr\space B=-7\\\Rarr\space 5x+3=\frac{5}{2}(2x+4)-7$$

$$\therefore\space \int\frac{5x+3}{\sqrt{x^2+4x+10}}dx\\=\int\frac{\frac{5}{2}(2x+4)-7}{\sqrt{x^2+4x+10}}dx\\=\frac{5}{2}\int\frac{2x+4}{\sqrt{x^2+4x+10}}dx\\-7\int\frac{1}{x^2+4x+10}dx\\\text{Let}\space\text{I}_1=\int\frac{2x+4}{\sqrt{x^2+4x+10}}dx\\\text{and}\space\text{I}_2=\int\frac{1}{\sqrt{x^2+4x+10}}dx\\\therefore\space\int\frac{5x+3}{\sqrt{x^2+4x+10}}dx$$

$$=\frac{5}{2}\text{I}_1-7\text{I}_2\space\text{...(i)}\\\text{Now,}\\\space\text{I}_1=\int\frac{2x+4}{\sqrt{x^2+4x+10}}dx\\\text{Let x}^2+4x+10=t\\\Rarr\space (2x+4)dx=dt\\\Rarr\space dx=\frac{dt}{2x+4}\\\therefore\space\text{I}_1=\int\frac{2x+4}{\sqrt{t}}×\frac{dt}{2x+4}\\=\int\frac{dt}{\sqrt{t}}=2\sqrt{t}+\text{C}_1\\=2\sqrt{x^2+4x+10}+\text{C}_1\space\text{...(ii)}$$

$$\text{and}\space\text{I}_2=\int\frac{1}{\sqrt{x^2+4x+10}}dx\\=\int\frac{1}{\sqrt{(x^2+4x+4)+6}}dx\\=\int\frac{1}{\sqrt{(x+2)^2+(\sqrt{6})^2}}\\=\text{log}|(x+2)+\sqrt{(x+2)^2+6}|+\text{C}_2\\\text{...(iii)}\\\bigg[\because\space\int\frac{dx}{\sqrt{x^2+a^2}}=\text{log}|x+\sqrt{x^2+a^2}|\bigg]\\=\text{log|x+2+}\sqrt{x^2+4x+10}|+\text{C}_2$$

On substituting the values of I1 and I2 from equations (ii) and (iii) in equation (i), we get

$$\int\frac{5x+3}{\sqrt{x^2+4x+10}}dx\\=\frac{5}{2}[2\sqrt{x^2+4x+10}]-7\space\text{log}\\|x+2+2\sqrt{x^2+4x+10}|+\text{C}\\\bigg[\because\space\frac{5}{2}\text{C}_1-7\text{C}_2=\text{C}\bigg]\\=5\sqrt{x^2+4x+10}\text{log}\\|x+2+\sqrt{x^2+4x+10}|+\text{C}$$

Direction (For Q. 24 and 25) : Choose the correct answer.

$$\textbf{24.}\space\int\frac{\textbf{1}}{\textbf{x}^\textbf{2}\textbf{+2x+2}}\textbf{dx}\space\textbf{equals}$$

(a) x tan–1 (x + 1) + C

(b) tan–1 (x + 1) + C

(c) (x + 1) tan–1 x + C

(d) tan–1 x + C

Sol. (b) tan–1 (x + 1) + C

$$\text{Let}\space\text{I}=\int\frac{1}{x^2+2x+2}dx\\=\int\frac{1}{(x+1)^2+1^2}dx\\\text{Let}\space x+1=t\\\Rarr\space dx=dt\\\therefore\space\text{I}=\int\frac{1}{t^2+1^2}dt\\=\frac{1}{1}\text{tan}^{\normalsize-1}\bigg(\frac{t}{1}\bigg)+\text{C}\\\bigg[\because\int\frac{dx}{a^2+x^2}=\frac{1}{a}\text{tan}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]\\=\text{tan}^{\normalsize-1}\bigg(\frac{x+1}{1}\bigg)+\text{C}$$

= tan–1 (x + 1) + C

$$\textbf{25.}\space\int\frac{\textbf{1}}{\sqrt{\textbf{9x-4x}^\textbf{2}}}\textbf{dx is equal}\\\textbf{(a)\space}\frac{\textbf{1}}{\textbf{9}}\textbf{sin}^{\normalsize-1}\bigg(\frac{\textbf{9x-8}}{\textbf{8}}\bigg)+\textbf{C}\\\textbf{(b)}\space\frac{\textbf{1}}{\textbf{2}}\textbf{sin}^{\normalsize-1}\bigg(\frac{\textbf{8x-9}}{\textbf{9}}\bigg)+\textbf{C}\\\textbf{(c)}\space\frac{\textbf{1}}{\textbf{3}}\textbf{sin}^{\normalsize-1}\bigg(\frac{\textbf{9x-8}}{\textbf{8}}\bigg)+\textbf{C}\\\textbf{(d)}\space\frac{\textbf{1}}{\textbf{2}}\space\textbf{sin}^{\normalsize-1}\bigg(\frac{\textbf{9x-8}}{\textbf{9}}\bigg)+\textbf{C}\\\textbf{Sol.}\space\text{(b)}\frac{1}{2}\text{sin}\bigg(\frac{8x-9}{a}\bigg)+\text{C}\\\int\frac{1}{\sqrt{9x-4x^2}}dx$$

$$\frac{\text{1}}{\sqrt{\text{4}}}\int\frac{1}{\sqrt{\frac{9}{4}x-x^2}}dx\\=\frac{1}{2}\int\frac{1}{\sqrt{-\bigg[x^2-\frac{9}{4}x+\bigg(\frac{9}{8}\bigg)^2\bigg]}+\bigg(\frac{9}{8}\bigg)^2}dx\\=\frac{1}{2}\int\frac{1}{\sqrt{\bigg(\frac{9}{8}\bigg)^2-\bigg(x-\frac{9}{8}\bigg)^2}}dx\\=\frac{1}{2}\text{sin}^{\normalsize-1}\bigg(\frac{x-\frac{9}{8}}{\frac{9}{8}}\bigg)+\text{C}\\\bigg[\because\space\int\frac{dx}{\sqrt{a^2-x^2}}=\text{sin}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]$$

$$=\frac{1}{2}\text{sin}^{\normalsize-1}\bigg(\frac{8x-9}{9}\bigg)+\text{C}$$