Question
Evaluate $\int_0^1 \frac{d x}{x^2+2 x+2}$.

Answer

Let
$I =\int_0^1 \frac{d x}{x^2+2 x+2}$
$ =\int_0^1 \frac{d x}{x^2+2 x+1+1}=\int_0^1 \frac{d x}{(x+1)^2+1}$
$ =\int_0^1 \frac{d x}{(x+1)^2+(1)^2}$
Let $ x+1=t$
$\therefore d x=d t$
If
$x=0$ then $t=1$
$x=1$ then $t=2$
So
$I =\int_1^2 \frac{d t}{t^2+1}=\left(\tan ^{-1} t\right)_1^2$
$ =\tan ^{-1} 2-\tan ^{-1} 1$
$ =\left(\tan ^{-1} 2-\frac{\pi}{4}\right)$

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