$\int {\underset{I}x^2 \underset{II}e^x} d x$

Solution:

$=x^2\left(\int e^x d x\right)-\int 2 x\left(\int e^x d x\right) d x$

$=x^2 e^x - 2 \int x e^x d x$

$=x^2 e^x - 2\left[x\left(\int e^x d x\right)-\int 1\left(\int e^x d x\right) d x\right]$

$=x^2 e^x-2\left[x e^x-\int e^x d x\right]$

$\int x^2 e^x=x^2 e^x-2 x e^x+2 e^x+C$

$\int \underset{I}x \underset{II}\sin 3 x d x$

Solution:

$= x \int \sin 3 x d x-\int 1 \cdot(\int\sin 3 x d x)d x$

$=x\left(\frac{-\cos 3 x}{3}\right)-\int\left(-\frac{\cos 3 x}{3}\right) d x$

$[\int f(a x+b) d x = \frac{F(ax+b)}{a}+c$

$\quad if \int f(x)dx = F(x)+c ]$

$=-\frac{x \cos 3 x}{3}+\frac{1}{3} \int \cos 3 x d x$

$=-\frac{x \cos 3 x}{3}+\frac{1}{3}\left[\frac{\sin 3 x}{3}\right]+c$

$\int x \sin 3 x=-x \frac{\cos 3 x}{3}+\frac{1}{9} \sin 3 x+c$

$\int \underset{I}f(x) \underset{II}g(x) d x$

$= \underset{I}f(x). \int \underset{II}g(x) d x-\int \underset{I}f^{\prime}(x) (\int \underset{II}g(x) d x) d x$

$I=\int {x} \log x d x$

Solution:

$I=\int \underset{I}{x} \underset{II}\log x d x$

$=x\left(\int \log x d x\right)- \int 1 \cdot\left(\int \log x\right) d x$

$=x {\int \log x d x}-{\int\left(\int \log x d x\right) d x}$

$\int \log x d x=\text { ? }$

$I=\int \underset{II}{x} \underset{I}\log x d x$

$=\log x \cdot \int x d x-$

$\int \frac{1}{x} \times\left(\int x d x\right) d x$

I=$\int \underset{II}{x} \underset{I}{\sin ^{-1} x} d x$

Solution:

$=\sin ^{-1} x \cdot \frac{x^2}{2}-\int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} d x$

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

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

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

$I_1=\int {\sqrt{1-x^2}}{d x},$

$I_2=\int \frac{1}{\sqrt{1-x^2}} d x$

$Put, x=\sin \theta$

$\Rightarrow \theta=\sin ^{-1} x$

$\Rightarrow \sqrt{1-x^2}=\sqrt{1-\sin ^2\theta} = \cos \theta ;$

$I_1 =\int {\cos \theta}.{\cos \theta d \theta}$

$[\cos 2 \theta=2 \cos ^2 \theta-1]$

$[\cos ^2 \theta=\frac{\cos 2 \theta-1}{2}]$

$I_1 =\int \cos ^2 \theta d\theta$

$=\frac{1}{2} \int(\cos 2 \theta-1) d\theta$

$=\frac{1}{2}(\frac{\sin 2 \theta}{2}-\theta)$

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

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

$I=\frac{x^2 \sin ^{-1} x}{2}-\frac{1}{2}.[\frac{x}{2} \sqrt{1-x^2}-\frac{1}{2} \sin ^{-1} x] -\frac{1}{2} \sin ^{-1} x+c$

Inverse trigonometric

Log function

Algebraic

Trigonometric

Exponential

$\int \underset{II}x \underset{I}{logx} dx$

$I=\int \log x d x$

Solution:

$=\int \underset{II}1 \cdot \underset{I}\log x d x$

$=\log x \cdot x-\int \frac{1}{x}.x d x$

$\int \log x d x=x \log x-x+c$

$I =\int \underset{I}x \underset{II}\log x d x$

Solution

$=x \int \log x d x-\int 1 \cdot\left(\int \log x d x\right) d x$

$=x(x \log x-x)-\int(x \log x-x) d x$

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

$I =x^2 \log x-\frac{x^2}{2}-I+C$

$I =\frac{x^2 \log x}{2}-\frac{x^2}{4}+C$

$I=\int \tan ^{-1} x d x$

Solution:

$=\int \underset{II}1 \cdot \underset{I}\tan ^{-1} x d x$

$=\tan ^{-1} x \cdot x-\int \frac{1}{1+x^2}. x d x$

$=x \tan ^{-1} x-\int \frac{x}{1+x^2} d x \quad$

[Put $1+x^2=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{d t}{2}$]

$=x \tan ^{-1} x-\frac{1}{2} \int \frac{d t}{t}$

$=x \tan ^{-1} x-\frac{1}{2} \log |t|+C$

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

$I_2=\int e^{m x} \sin n x d x$

Solution:

$I_2=\int \underset{II}{e^{mx}} \underset{I}{\sin n x} d x$

$=\sin n x \cdot \frac{e^{m x}}{m}-\int n \cos n x \frac{e^{m x}}{m} d x$

$=\frac{e^{m x} \sin nx}{m} - \frac{n}{m}\left[\int \underset{II}e^{m x} \underset{I}\cos n x d x\right]$

$=\frac{e^{m x} \sin n x}{m}-\frac{n}{m} \left[\cos n x \cdot \frac{e^{mx}}{m}-\int n(-\sin nx).\frac{e^{mx}}{m} d x\right]$

$I_2=\frac{e^{m x} \sin n x}{m}-\frac{n}{m^2} e^{m x} \cos n x \ -\frac{n^2}{m^2} {\int e^{m x} \sin n x d x}$

[$\because I_2 = \int e^{m x} \sin n x d x$ ]

$\Rightarrow I_2\left(1+\frac{n^2}{m^2}\right)=\frac{m e^{m x} \sin n x-n e^{m x} \cos n x}{m^2}$

$\Rightarrow I_2=\int e^{m x} \sin n x d x=\frac{1}{m^2+n^2}[m \sin n x-n \cos nx] e^{m x}$

1.

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

$=\int \underset{II}{1} \cdot \underset{I}{\sqrt{x^2-a^2}} d x$

$=\sqrt{x^2-a^2}. x-\int \frac{2 x}{2 \sqrt{x^2-a^2}} \times x dx$

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

$I= x \sqrt{x^2-a^2}-\int \sqrt{x^2-a^2} d x -a^2 \int \frac{1}{\sqrt{x^2-a^2}} d x$

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

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

2. $\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+c$

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

$I= \int \sqrt{5-(x+2)^2} d x$

[$Put, x+2=t \Rightarrow dx=dt$ ]

$= \int \sqrt{5-t^2} d t$

$= \frac{t}{2} \sqrt{5-t^2}+\frac{5}{2} \sin ^{-1} \frac{t}{\sqrt{5}}+C$

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