Answer
\[I=\tan^{-1}(x+1)+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int\frac{1}{x^2+2x+2}dx\]
\[I=\int\frac{1}{(x+1)^2+1}dx\]
Put $\;t=x+1$ ___(1)
$\Rightarrow dt=dx$
\[I=\int\frac{1}{t^2+1}dt\]
We will use the formula \[\int\frac{1}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}\;\;\;...(2)\]
By using (2)
\[I=\tan^{-1}\frac{t}{1}+C\]
Where $C$ is constant of integration
By using (1)
\[I=\tan^{-1}(x+1)+C\]
Hence
\[I=\tan^{-1}(x+1)+C\;\;.\]
