What is the integral from 0 to 2 of x/((x^2-2x+2)) ?

The integral from 0 to 2 of x/((x^2-2x+2)) is pi/2

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integral from 0 to 2 of x/((x^2-2x+2))

\int_{0}^{2} \frac{x}{( x ^{2} - 2 x + 2 )} d x

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

\int_{0}^{2} \frac{x}{x ^{2} - 2 x + 2} d x

Complete the square

x^{2} - 2 x + 2 : (x - 1)^{2} + 1

= \int_{0}^{2} \frac{x}{( x - 1 ) ^{2} + 1} d x

Apply u-substitution

= \int_{- 1}^{1} \frac{u + 1}{u ^{2} + 1} d u

Expand

\frac{u + 1}{u ^{2} + 1} : \frac{u}{u ^{2} + 1} + \frac{1}{u ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{1} \frac{u}{u ^{2} + 1} d u + \int_{- 1}^{1} \frac{1}{u ^{2} + 1} d u

\int_{- 1}^{1} \frac{u}{u ^{2} + 1} d u = 0

\int_{- 1}^{1} \frac{1}{u ^{2} + 1} d u = \frac{π}{2}

= 0 + \frac{π}{2}

Simplify

= \frac{π}{2}

What is the integral from 0 to 2 of x/((x^2-2x+2)) ?
The integral from 0 to 2 of x/((x^2-2x+2)) is pi/2