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

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

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Answer 1 · 2017-02-04T15:32:02

$\frac{π}{2}$ $pi/2$

Explanation:

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

$\int_{0}^{2} \frac{1}{1 (x^{2} - 2 x + 1 - 1) + 2} d x$ $int_0^2 1/(1(x^2 - 2x + 1 - 1) + 2)dx$

$\int_{0}^{2} \frac{1}{1 (x^{2} - 2 x + 1) - 1 + 2} d x$ $int_0^2 1/(1(x^2 - 2x + 1) - 1 + 2)dx$

$\int_{0}^{2} \frac{1}{{(x - 1)}^{2} + 1} d x$ $int_0^2 1/((x -1)^2 + 1)dx$

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

$\int_{- 1}^{1} \frac{1}{u^{2} + 1} d u$ $int_-1^1 1/(u^2 + 1) du$

This is a standard integral.

$\int_{- 1}^{1} arctan (u)$ $int_-1^1 arctan(u)$

$\int_{0}^{2} arctan (x - 1)$ $int_0^2 arctan(x- 1)$

Evaluate using the 2nd fundamental theorem of calculus.

$arctan (2 - 1) - arctan (0 - 1) = arctan (1) - arctan (- 1) = \frac{π}{4} - (- \frac{π}{4}) = \frac{π}{2}$ $arctan(2 - 1) - arctan(0 - 1) = arctan(1) - arctan(-1) = pi/4 - (-pi/4) = pi/2$

Hopefully this helps!

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

1 Answer

Explanation:

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