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

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Answer 1 · 2017-02-04T15:32:02

$pi/2$

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

$int_0^2 1/(1(x^2 - 2x + 1 - 1) + 2)dx$

$int_0^2 1/(1(x^2 - 2x + 1) - 1 + 2)dx$

$int_0^2 1/((x -1)^2 + 1)dx$

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

$int_-1^1 1/(u^2 + 1) du$

This is a standard integral.

$int_-1^1 arctan(u)$

$int_0^2 arctan(x- 1)$

Evaluate using the 2nd fundamental theorem of calculus.

$arctan(2 - 1) - arctan(0 - 1) = arctan(1) - arctan(-1) = pi/4 - (-pi/4) = pi/2$

Hopefully this helps!

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