Evaluate the following integrals:

We can write $x^{2}+2 x+1+1=(x+1)^{2}+1$

$\Rightarrow \frac{1 \cdot d x}{(x+1)(x+1)^{2}+1}$

Assume $x+1=$ tant

$\Rightarrow d x=\sec ^{2} t \cdot d x$

$\Rightarrow \int \frac{\sec ^{2} t d t}{\tan t \cdot \tan ^{2} t+1}$

$\Rightarrow \tan ^{2} t+1=\sec ^{2} t$

$\Rightarrow \int \frac{d t}{\tan t}$

$\Rightarrow \frac{\cos t}{\sin t} d t$

$\Rightarrow \log |\sin t|+c$

$\Rightarrow \sin t=\frac{\tan t}{\sec ^{2} t}$

But tant $=x+1$

$\Rightarrow \sin t=\frac{x+1}{(1+x)^{2}+1}$

The final answer is

$\Rightarrow \log \sin \left|\frac{x+1}{x^{2}+2 x+2}\right|+c$