• $I =\int \frac{d x}{x^2+a^2}$

• Solution

• $x=a \tan t \quad \Rightarrow d x=a \sec ^2 t d t$

• $x^2+a^2 =a^2 \tan ^2 t+a^2= a^2(\tan ^2 t+1)=a^2 \sec ^2 t$

• $=\int \frac{a \sec ^2 t d t}{a^2 \operatorname{sec}^2 t}$

• $=\frac{1}{a} \int d t=\frac{1}{a} t+c=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$

• $\int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$

• $I=\int \frac{d x}{\sqrt{x^2+a^2}}$

• solution:

• $x=a \tan t \Rightarrow d x=a \sec ^2 t d t$

• $t=\tan ^{-1} \frac{x}{a}, \quad x^2+a^2=a^2 \sec ^2 t$

• $I=\int \frac{a \sec ^2 t d t}{\sqrt{a^2 \sec ^2 t}}=\int \sec t d t$

• [$\int \sec t d t =\log |\sec t+\tan t|+c$ ]

• $=\log |\sec t+\tan t|+c$

• $[ sec ^2 t=1+\tan ^2 t ]$

• $[ \sec t =\sqrt{1+\tan ^2 t}]$

• $=\log |\sqrt{1+\tan ^2 t}+\tan t| + c$

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

• $=\log \left|x+\sqrt{x^2+a^2}\right|-\log |a|+c$

• $\int \frac{d x}{\sqrt{x^2+a^2}}=\log \left|x+\sqrt{x^2+a^2}\right|+C$

• $I =\int \frac{d x}{x^2-a^2}$

• Solution:

• $[ A^2-B^2=(A-B)(A+B) ]$

• $=\int \frac{d x}{(x-a)(x+a)}$

• $\{\because \frac{1}{(x-a)(x+a)}=\frac{1}{2 a}\left[\frac{2 a}{(x+a)(x-a)}\right] =\frac{1}{2a}\left[\frac{(x+a)-(x-a)}{(x+a)(x-a)}\right] =\frac{1}{2 a}\left[\frac{1}{x-a}-\frac{1}{x+a}\right]\}$

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

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

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

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

• $\int \frac{d x}{a^2-x^2}$

• solution:

• $=\int \frac{1}{2 a}\left[\frac{(a-x)+(a+x)}{(a-x)(a+x)}\right] d x$

• $=\frac{1}{2 a}\left[\int \frac{d x}{a+x}+\int \frac{d x}{a-x}\right] \quad\left[\int \frac{d x}{a-x}=?\right]$

• $=\frac{1}{2 a}\left[\log |a+x|+\frac{\log |a-x|}{-1}\right]+c$

• $=\frac{1}{2 a}[\log |a+x|-\log |a-x|]+c$

• $\int \frac{d x}{a^2-x^2} =\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C$

• $I=\int \frac{3 x^2 d x}{x^6+4}$

• Solution:

• $=\int \frac{3 x^2 d x}{\left(x^3\right)^2+2^2} [\quad x^3=t \rightarrow\quad 3 x^2 d x=d t ]$

• $=\int \frac{d t}{t^2+2^2}=\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+C$

• $[\because \int \frac{d x}{x^2+a^2}=\frac{1}{a}. \tan ^{-1} \frac{x}{a}+c ]$

• $=\frac{1}{2} \tan ^{-1} \frac{x^3}{2}+c$

.

• $\int \frac{x^2 d x}{1-x^6}$

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

• $[ x^3=t ]$

• $[ 3 x^2 d x=d t ]$

• $[ x^2 d x=\frac{d t}{3} ]$

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

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

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

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

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

• $I =\int \frac{d x}{\sqrt{a^2-x^2}}$

• Solution

• $x =a \sin t, \quad d x=a \cos t d t$

• $I =\int \frac{a \cos t d t}{\sqrt{a^2\left(1-\sin ^2 t\right)}}=\int d t$

• $=t+C$

• $I =\sin ^{-1} \frac{x}{a}+C$

• $I =\int \frac{d x}{\sqrt{x^2-a^2}}$

• Solution

• $x=a \sec t \rightarrow d x=a \sec t \tan t d t$

• $x^2-a^2=a^2 \sec ^2 t-a^2$

• $=a^2\left(\sec ^2 t-1\right)$

• $=a^2 \tan ^2 t$

• $I=\int \frac{a \sec t \cdot \operatorname{tan} t \cdot d t}{a \tan t}=\int \sec t d t$

• $=\log |\sec t+\tan t|+C$

• [x = $a \sec t$]

• $[ \ sec t = \frac{x}{a} ]$

• $[ tant = \sqrt{\sec ^2 t-1} =\sqrt{\frac{x^2}{a^2} -1} ]$

• $=\log \left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right|+c$

• $\int \frac{d x}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+c$

• $I =\int \frac{d x}{ax^2+b x+c}$

• Solution:

• $a x^2+b x+c =a\left(x^2+\frac{b}{a} x+\frac{c}{a}\right)$

• $=a(\underbrace{x^2+2 \times \frac{b}{2 a} \times x+\frac{b^2}{4 a^2}}-\frac{b^2}{4 a^2}+\frac{c}{a})$

• $=a\left[\left(x+\frac{b}{2 a}\right)^2+\left(\frac{c}{a}-\frac{b^2}{4 a^2}\right)\right]$

• $x+\frac{b}{2 a}=X, \quad \frac{c}{a}-\frac{b^2}{4 a^2}= \pm k^2, d x=d X$

• $I=\frac{1}{a} \int \frac{d X}{X^2 \pm k^2}$

• $I=\int \frac{d x}{9 x^2+6 x+5}$

• Solution:

• $9 x^2+6 x+5 =9\left(x^2+\frac{6}{9} x+\frac{5}{9}\right)$

• $=9\left(x^2+\frac{2}{3} x+\frac{5}{9}\right)$

• $=9\left(x^2+2 \times \frac{1}{3} \times{x} +\frac{1}{9}-\frac{1}{9}+\frac{5}{9}\right)$

• $=9\left(\left(x+\frac{1}{3}\right)^2+\left(\frac{2}{3}\right)^2\right)$

• $I =\int \frac{d x}{9\left((x+1 / 3)^2+(2 / 3)^2\right)}=\frac{1}{9} \int \frac{d X}{X^2+(2 / 3)^2}$

• Solution:

• (Put $x+ \frac{1}{3} = X$)

• $=\frac{1}{9} \times \frac{1}{2 / 3} \tan ^{-1} \frac{X}{2 / 3}+C$

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

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

.

• $I=\int \frac{d x}{9 x^2+6 x+5}$

• Solution:

• $9 x^2+6 x+5=(3 x)^2+2(3 x)+1+4=(3 x+1)^2+4$

• $I =\int \frac{d x}{(3 x+1)^2+4}$

• [ 3 x+1=t ]

• [ 3 d x=d t ]

• $=\frac{1}{3} \int \frac{d t}{t^2+2^2}=\frac{1}{3} \cdot \frac{1}{2} \tan ^{-1}\frac{t}{2}+C$

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

• $\int \frac{d x}{\sqrt{a x^2+b x+c}}$

• Solution:

• $a x^2+b x+c \sim x^2 \pm k^2$

• $\sim k^2-x^2$

• Write $a x^2+b x+c$ into either of the form $x^2 \pm k^2$(if a is positive) , or ,$k^2-x^2$(if a is negative)

• $\int \frac{d x}{\sqrt{x^2+2 x+2}}$

• Solution:

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

• [x+1=t]

• [d x=dt]

• $=\int \frac{d t}{\sqrt{t^2+1}}$

• $=\log \left|t+\sqrt{t^2+1}\right|+C$

• $[ \because \int \frac{d x}{\sqrt{x^2+a^2}}=\log \mid x+ \sqrt{(x^2 + a^2)}+C ]$

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

• $=\log \left|x+1+\sqrt{x^2+2 x+2}\right|+c$

• $I_1 =\int \frac{p x+q}{a x^2+b x+c} d x$

• $I_2 =\int \frac{p x+q}{\sqrt{a x^2+b x+c}} d x$

• Solution:

• $p x+q=A \frac{d}{d x}\left(a x^2+b x+c\right)+B$

• $p x+q=A(2 a x+b)+B$

• $p=2 a A, q=A b+B$

• $A=? \quad B=?$

• $I_1 =A \int \frac{\frac{d}{d x}\left(a x^2+b x+c\right)}{a x^2+b x+c} d x+B \int \frac{1}{a x^2+b x+c}dx$

• $\rightarrow \text { can be evaluated. }$

• $I_1 =A \int \frac{\frac{d}{d x}\left(a x^2+b x+c\right)}{a x^2+b x+c} d x+B \int \frac{1}{a x^2+b x+c}$

• $\rightarrow \text { can be evaluated. }$

• $I =\int \frac{6 x-2}{3 x^2+2 x-1} d x$

• $6 x-2=A \frac{d}{d x}\left(3 x^2+2 x-1\right)+B$

• $6 x-2=A(6 x+2)+B \Rightarrow A=1, B=-4$

• $I=\int \frac{1 \cdot \frac{d}{d x}\left(3 x^2+2 x-1\right)}{3 x^2+2 x-1} d x+(-4) \int \frac{1}{3 x^2+2 x-1} d x$

• [Put $3 x^2+2 x-1 = t]$

• $=\int \frac{d t}{t}-4 \times \frac{1}{3} \int \frac{d x}{x^2+\frac{2}{3} x-1}$

• $[ x^2+\frac{2}{3} x-1 =\left(x+\frac{1}{3}\right)^2-\frac{4}{9} ]$

• $=\log t+C_1-\frac{4}{3} \int \frac{d x}{(x+1 / 3)^2-(2 / 3)^2}$

• $=\log \left|3 x^2+2 x-1\right|-\frac{4}{3} \cdot \frac{1}{2 . \frac{2}{3}}. \log \left|\frac{x+1 / 3-2 / 3}{x+1 / 3+2 / 3}\right|+C_1+C_2$

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