Evaluate the following integrals:
$\int \frac{1}{(x+1)\left(x^{2}+2 x+2\right)} d x$
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
$\Rightarrow \log \sin \left|\frac{x+1}{x^{2}+2 x+2}\right|+c$
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