Consider that, $$I=\int \frac{(1+\cos x)}{(x+\sin x)} d x$$

Let $$x+\sin x=t \Rightarrow(1+\cos x) d x=d t$$

$$\begin{aligned} \therefore\quad I & =\int_t^1 d t=\log |t|+C \\ & =\log |(x+\sin x)|+C \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int \frac{d x}{1+\cos x}=\int \frac{d x}{1+2 \cos ^2 \frac{x}{2}-1} \\ & =\frac{1}{2} \int \frac{1}{\cos ^2 \frac{x}{2}} d x=\frac{1}{2} \int \sec ^2 \frac{x}{2} d x \\ & =\frac{1}{2} \cdot \tan \frac{x}{2} \cdot 2+C=\tan \frac{x}{2}+C \quad\left[\because \int \sec ^2 x d x=\tan x\right] \end{aligned}$$

Let $$I=\int \tan ^2 x \sec ^4 x d x$$

$$\begin{array}{lrl} \text { Put } & \tan x & =t \Rightarrow \sec ^2 x d x=d t \\ \therefore & I & =\int t^2\left(1+t^2\right) d t=\int\left(t^2+t^4\right) d t \end{array}$$

$=\frac{t^3}{3}+\frac{t^5}{5}+C=\frac{\tan ^5 x}{5}+\frac{\tan ^3 x}{3}+C$

$$\begin{aligned} \text { Let } \quad I & =\int \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x=\int \frac{(\sin x+\cos x)}{\sqrt{\sin ^2 x+\cos ^2 x+2 \sin x \cos x}} d x \\ & =\int \frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^2}} d x=\int 1 d x=x+C \end{aligned} $$

$$\begin{aligned} \text{Let}\quad I & =\int \sqrt{1+\sin x} d x \\ & =\int \sqrt{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}} d x \quad\left[\because \sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}=1\right] \\ & =\int \sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2} d x=\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x \\ & =-\cos \frac{x}{2} \cdot 2+\sin \frac{x}{2} \cdot 2+C=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}+C \end{aligned}$$