Answer
$\frac{\pi}{2}$
Work Step by Step
$\int_{0}^{2}\frac{dx}{x^{2}-2x+2}$
To solve this integral, we must manipulate the denominator by completing the square:
$x^{2}-2x+2$
$x^{2}-2x+1+2-1$
$(x-1)^{2}+1$
$\int_{0}^{2}\frac{dx}{(x-1)^{2}+1}$
Now with the new manipulated denominator, we can solve the integral using the integral of arctan:
$\int\frac{1}{a^{2}+u^{2}}dx=\frac{1}{a}arctan(\frac{u}{a})+C$
$u=x-1$
$du=dx$
$a=1$
$[\frac{1}{1}arctan(\frac{x-1}{1})]_{0}^{2}$
$[arctan(2-1)-arctan(0-1)]$
$[arctan1-arctan(-1)]$
$[\frac{\pi}{4}+\frac{\pi}{4}]$
$\frac{\pi}{2}$
Note that $arctan(-1)$ is equal to $\frac{-\pi}{4}$ because the bounds of the domain of $arctan$ are limited to $[\frac{-\pi}{2},\frac{\pi}{2}]$.
