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!