Let $I=\int e^{\tan ^{-1} x}\left(\frac{1+x+x^2}{1+x^2}\right) d x$

$$\begin{aligned} & =\int e^{\tan ^{-1} x}\left(\frac{1+x^2}{1+x^2}+\frac{x}{1+x^2}\right) d x \\ & =\int e^{\tan ^{-1} x} d x+\int \frac{x e^{\tan ^{-1} x}}{1+x^2} d x \end{aligned}$$

$I=I_1+I_2\quad\text{.... (i)}$

Now, $I_2=\int \frac{x \mathrm{e}^{\tan ^{-1} x}}{1+x^2} d x$

Put $\tan ^{-1} x=t \Rightarrow x=\tan t$

$\Rightarrow \quad \frac{1}{1+x^2} d x=d t$

$\therefore\quad I=\int_{\mathrm{I}}^{\tan } t \cdot e_{\mathrm{II}}^t d t$

$$\begin{aligned} & =\tan t \cdot e^t-\int \sec ^2 t \cdot e^t d t+C \\ & =\tan t \cdot e^t-\int\left(1+\tan ^2 t\right) e^t d t+C \quad\left[\because \sec ^2 \theta=1+\tan ^2 \theta\right] \end{aligned}$$

$$\begin{aligned} I_2 & =\tan t \cdot e^t-\int\left(1+x^2\right) \frac{e^{\tan ^{-1} x}}{1+x^2} d x+C \\ I_2 & =\tan t \cdot e^t-\int e^{\tan ^{-1} x} d x+C \\ \therefore\quad I & =\int e^{\tan ^{-1} x} d x+\tan t \cdot e^t-\int e^{\tan ^{-1} x} d x+C \\ & =\tan t \cdot e^t+C \\ & =x e^{\tan ^{-1} x}+C \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x \\ \text{Put}\quad x & =a \tan ^2 \theta \\ \Rightarrow\quad d x & =2 a \tan \theta \sec ^2 \theta d \theta \\ \therefore\quad I & =\int \sin ^{-1} \sqrt{\frac{a \tan ^2 \theta}{a+a \tan ^2 \theta}}\left(2 a \tan \theta \cdot \sec ^2 \theta\right) d \theta \\ & =2 a \int \sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right) \tan \theta \cdot \sec ^2 \theta d \theta \\ & =2 a \int \sin ^{-1}(\sin \theta) \tan \theta \cdot \sec ^2 \theta d \theta \end{aligned}$$

$$\begin{aligned} & =2 a \int_{\text {I }}^\theta \theta \cdot \tan \theta \sec ^2 \theta d \theta \\ & =2 a\left[\theta \cdot \int \tan \theta \cdot \sec ^2 \theta d \theta-\int\left(\frac{d}{d \theta} \theta \cdot \int \tan \theta \cdot \sec ^2 \theta d \theta\right) d \theta\right] \end{aligned}$$

$\left[\begin{array}{lrl}\text { Put } & \tan \theta & =t \\ \Rightarrow & \sec \theta \cdot \tan \theta \cdot d \theta & =d t \\ \Rightarrow & \int \tan \theta \sec ^2 \theta d \theta & =\int t d t\end{array}\right]$

$$\begin{aligned} & =2 a\left[\theta \cdot \frac{\tan ^2 \theta}{2}-\int \frac{\tan ^2 \theta}{2} d \theta\right] \\ & =a \theta \tan ^2 \theta-a \int\left(\sec ^2 \theta-1\right) d \theta \\ & =a \theta \cdot \tan ^2 \theta-a \tan \theta+a \theta+C \\ & =a\left[\frac{x}{a} \tan ^{-1} \sqrt{\frac{x}{a}}+\tan ^{-1} \sqrt{\frac{x}{a}}\right]+C \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{5 / 2}} d x \\ & =\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^2 \sqrt{1+\cos x}} d x \\ & =\int_{\pi 3}^{\pi 2} \frac{1}{\left(1-\cos ^2 x\right)} d x=\int_{\pi 3}^{\pi 2} \frac{1}{\sin ^2 x} d x \\ & =\int_{\pi 3}^{\pi 2} \operatorname{cosec}^2 x d x=[-\cot x]_{\pi / 3}^{\pi / 2} \\ & =-\left[\cot \frac{\pi}{2}-\cot \frac{\pi}{3}\right]=-\left[0-\frac{1}{\sqrt{3}}\right]=+\frac{1}{\sqrt{3}} \end{aligned}$$

Alternate Method

Let $I=\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{5 / 2}} d x=\int_{\pi / 3}^{\pi / 2} \frac{\left(2 \cos ^2 \frac{x}{2}\right)^{1 / 2}}{\left(2 \sin ^2 \frac{x}{2}\right)^{5 / 2}} d x$

$=\frac{\sqrt{2}}{4 \sqrt{2}} \int_{\pi / 3}^{\pi / 2} \frac{\cos \left(\frac{x}{2}\right)}{\sin ^5\left(\frac{x}{2}\right)} d x=\frac{1}{4} \int_{\pi / 3}^{\pi / 2} \frac{\cos \left(\frac{x}{2}\right)}{\sin ^5\left(\frac{x}{2}\right)} d x$

$$\begin{array}{lr} \text { Put } & \sin \frac{x}{2}=t \\ \Rightarrow & \cos \frac{x}{2} \cdot \frac{1}{2} d x=d t \\ \Rightarrow & \cos \frac{x}{2} d x=2 d t \end{array}$$

As $\quad x \rightarrow \frac{\pi}{3}$, then $t \rightarrow \frac{1}{2}$

and $x \rightarrow \frac{\pi}{2}$, then $t \rightarrow \frac{1}{\sqrt{2}}$

$$\begin{aligned} \therefore\quad I & =\frac{2}{4} \int_{1 / 2}^{1 \sqrt{2}} \frac{d t}{t^5}=\frac{1}{2}\left[\frac{t^{-5+1}}{-5+1}\right]_{1 / 2}^{1 / \sqrt{2}} \\ & =-\frac{1}{8}\left[\frac{1}{\left(\frac{1}{\sqrt{2}}\right)^4}-\frac{1}{\left(\frac{1}{2}\right)^4}\right]^8 \\ & =-\frac{1}{8}(4-16)=\frac{12}{8}=\frac{3}{2} \end{aligned}$$

Let $$I = \int {\mathop {{e^{ - 3x}}}\limits_{II} {{\cos }^3}x\,\mathop {dx}\limits_I } $$

$$\begin{aligned} & =\cos ^3 x \int e^{-3 x} d x-\int\left(\frac{d}{d x} \cos ^3 x \int e^{-3 x} d x\right) d x \\ & =\cos ^3 x \cdot \frac{e^{-3 x}}{-3}-\int\left(-3 \cos ^2 x\right) \sin x \cdot \frac{e^{-3 x}}{-3} d x \\ & =-\frac{1}{3} \cos ^3 x e^{-3 x}-\int \cos ^2 x \sin x e^{-3 x} d x \\ & =-\frac{1}{3} \cos ^3 x e^{-3 x}-\int\left(1-\sin ^2 x\right) \sin x e^{-3 x} d x \end{aligned}$$

$=-\frac{1}{3} \cos ^3 x e^{-3 x}-\int \sin x e^{-3 x} d x+$ $$\int {\mathop {{{\sin }^3}}\limits_I x\,{e^{ - 3x}}\mathop {dx}\limits_{II} } $$

$$\begin{aligned} & =-\frac{1}{3} \cos ^3 x e^{-3 x}-\int \sin x e^{-3 x} d x+\sin ^3 x \cdot \frac{e^{-3 x}}{-3}-\int 3 \sin ^2 x \cos x \cdot \frac{e^{-3 x}}{-3} d x \\ & =-\frac{1}{3} \cos ^3 x e^{-3 x}-\int \sin x e^{-3 x} d x-\frac{1}{3} \sin ^3 x e^{-3 x}+\int\left(1-\cos ^2 x\right) \cos x e^{-3 x} d x \\ I & =-\frac{1}{3} \cos ^3 x e^{-3 x}-\int \sin x e^{-3 x}-\frac{1}{3} \sin ^3 x e^{-3 x}+\int \cos x e^{-3 x} d x-\int \cos ^3 x e^{-3 x} d x \\ 2 I & =\frac{e^{-3 x}}{3}\left[\cos ^3 x+\sin ^3 x\right]-\left[\sin x \cdot \frac{e^{-3 x}}{-3}-\int \cos x \cdot \frac{e^{-3 x}}{-3} d x\right]+\int \cos x e^{-3 x} d x \\ 2 I & =\frac{e^{-3 x}}{-3}\left[\cos ^3 x+\sin ^3 x\right]+\frac{1}{3} \sin x \cdot e^{-3 x}-\frac{1}{3} \int \cos x \cdot e^{-3 x} d x+\int \cos x e^{-3 x} d x \\ 2 I & =\frac{e^{-3 x}}{-3}\left[\cos ^3 x+\sin ^3 x\right]+\frac{1}{3} \sin x e^{-3 x}+\frac{2}{3} \int \cos x e^{-3 x} d x \end{aligned}$$

Now, Let $${I_1} = \int {\mathop {\cos }\limits_I x\,{e^{ - 3x}}\mathop {dx}\limits_{II} } $$

$$\begin{aligned} & I_1=\cos x \cdot \frac{e^{-3 x}}{-3}-\int(-\sin x) \cdot \frac{e^{-3 x}}{-3} d x \\ & I_1=\frac{-1}{3} \cos x \cdot e^{-3 x}-\frac{1}{3} \int \sin x \cdot e^{-3 x} d x \end{aligned}$$

$$\begin{aligned} & =-\frac{1}{3} \cos x \cdot e^{-3 x}-\frac{1}{3}\left[\sin x \cdot \frac{e^{-3 x}}{-3}-\int \cos x \cdot \frac{e^{-3 x}}{-3} d x\right] \\ & =-\frac{1}{3} \cos x \cdot e^{-3 x}+\frac{1}{9} \sin x \cdot e^{-3 x}-\frac{1}{9} \int \cos x \cdot e^{-3 x} d x \end{aligned}$$

$$\begin{aligned} I_1+\frac{1}{9} I_1 & =-\frac{1}{3} e^{-3 x} \cdot \cos x+\frac{1}{9} \sin x \cdot e^{-3 x} \\ \left(\frac{10}{9}\right) I_1 & =-\frac{1}{3} e^{-3 x} \cdot \cos x+\frac{1}{9} \sin x \cdot e^{-3 x} \\ I_1 & =\frac{-3}{10} e^{-3 x} \cdot \cos x+\frac{1}{10} e^{-3 x} \sin x \end{aligned}$$

$\begin{aligned} 2 I=-\frac{1}{3} e^{-3 x}\left[\sin ^3 x+\cos ^3 x\right]+\frac{1}{3} \sin x \cdot e^{-3 x}-\frac{3}{10} e^{-3 x} & \cdot \cos x \\ & +\frac{1}{10} e^{-3 x} \cdot \sin x+C\end{aligned}$

$$\begin{aligned} \therefore\quad & I=-\frac{1}{6} e^{-3 x}\left[\sin ^3 x+\cos ^3 x\right]+\frac{13}{30} e^{-3 x} \cdot \sin x-\frac{3}{10} e^{-3 x} \cdot \cos x+C \\ & \qquad\left[\begin{array}{l} \because \sin 3 x=3 \sin x-4 \sin ^3 x \\ \text { and } \cos 3 x=4 \cos ^3 x-3 \cos x \end{array}\right] \\ & =\frac{e^{-3 x}}{24}[\sin 3 x-\cos 3 x]+\frac{3 e^{-3 x}}{40}[\sin x-3 \cos x]+C \end{aligned}$$

Let $$I=\int \sqrt{\tan x} d x$$

Put $\tan x=t^2 \Rightarrow \sec ^2 x d x=2 t d t$

$$\begin{aligned} \therefore\quad I & =\int t \cdot \frac{2 t}{\sec ^2 x} d t=2 \int \frac{t^2}{1+t^4} d t \\ & =\int \frac{\left(t^2+1\right)+\left(t^2-1\right)}{\left(1+t^4\right)} d t \\ & =\int \frac{t^2+1}{1+t^4} d t+\int \frac{t^2-1}{1+t^4} d t \\ & =\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} d t+\int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}} d t \\ & =\int \frac{1-\left(-\frac{1}{t^2}\right) d t}{\left(t-\frac{1}{t}\right)^2+2}+\int \frac{1+\left(-\frac{1}{t^2}\right)}{\left(t+\frac{1}{t}\right)^2-2} d t \end{aligned}$$

Put $$u=t-\frac{1}{t} \Rightarrow d u=\left(1+\frac{1}{t^2}\right) d t$$

and $$v=t+\frac{1}{t} \Rightarrow d v=\left(1-\frac{1}{t^2}\right) d t$$

$$\begin{aligned} \therefore\quad I & =\int \frac{d u}{u^2+(\sqrt{2})^2}+\int \frac{d v}{v^2-(\sqrt{2})^2} \\ & =\frac{1}{\sqrt{2}} \tan ^{-1} \frac{u}{\sqrt{2}}+\frac{1}{2 \sqrt{2}} \log \left|\frac{v-\sqrt{2}}{v+\sqrt{2}}\right|+C \\ & =\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+\frac{1}{2 \sqrt{2}} \log \left|\frac{\tan x-\sqrt{2 \tan x}+1}{\tan x+\sqrt{2 \tan x}+1}\right|+C \end{aligned}$$