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

Answer

Let
$
\begin{aligned}
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}
\end{aligned}
$
Let $\quad x+1=t \therefore d x=d t$
If
$
\begin{array}{l}
x=0 \text { then } t=1 \\
x=1 \text { then } t=2
\end{array}
$
So
$
\begin{aligned}
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) \text { }
\end{aligned}
$

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