# How do you evaluate int dx/(x^2-2x+2) from [0, 2]?

Feb 4, 2017

$\frac{\pi}{2}$

#### Explanation:

The denominator cannot be factored, so complete the square and see what you can do in the way of u-substitution.

${\int}_{0}^{2} \frac{1}{1 \left({x}^{2} - 2 x + 1 - 1\right) + 2} \mathrm{dx}$

${\int}_{0}^{2} \frac{1}{1 \left({x}^{2} - 2 x + 1\right) - 1 + 2} \mathrm{dx}$

${\int}_{0}^{2} \frac{1}{{\left(x - 1\right)}^{2} + 1} \mathrm{dx}$

Now let $u = x - 1$. Then $\mathrm{du} = \mathrm{dx}$. We also adjust our bounds of integration accordingly.

${\int}_{-} {1}^{1} \frac{1}{{u}^{2} + 1} \mathrm{du}$

This is a standard integral.

${\int}_{-} {1}^{1} \arctan \left(u\right)$

${\int}_{0}^{2} \arctan \left(x - 1\right)$

Evaluate using the 2nd fundamental theorem of calculus.

$\arctan \left(2 - 1\right) - \arctan \left(0 - 1\right) = \arctan \left(1\right) - \arctan \left(- 1\right) = \frac{\pi}{4} - \left(- \frac{\pi}{4}\right) = \frac{\pi}{2}$

Hopefully this helps!