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

The integral from 0 to 2 of x/((x^2-2x+2)^2) is (2pi+4)/8

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

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

\frac{2 π + 4}{8}

Decimal

1.28539 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int_{- 1}^{1} \frac{u + 1}{( u ^{2} + 1 ) ^{2}} d u

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

Apply Integration By Parts

: [\frac{1}{2} (u + 1) (arctan (u) + \frac{u}{1 + u ^{2}}) - \int \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}) d u]_{- 1}^{1}

= [\frac{1}{2} (u + 1) (arctan (u) + \frac{u}{1 + u ^{2}}) - \int \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}) d u]_{- 1}^{1}

\int \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}) d u = \frac{1}{2} (u (arctan (u) + \frac{u}{1 + u ^{2}}) + \frac{1}{u ^{2} + 1})

= [\frac{1}{2} (u + 1) (arctan (u) + \frac{u}{1 + u ^{2}}) - \frac{1}{2} (u (arctan (u) + \frac{u}{1 + u ^{2}}) + \frac{1}{u ^{2} + 1})]_{- 1}^{1}

Join

\frac{1}{2} (u + 1) (arctan (u) + \frac{u}{1 + u ^{2}}) - \frac{1}{2} (u (arctan (u) + \frac{u}{1 + u ^{2}}) + \frac{1}{u ^{2} + 1}) : \frac{u ^{2} arctan ( u ) + u + arctan ( u ) - 1}{2 ( u ^{2} + 1 )}

= [\frac{u ^{2} arctan ( u ) + u + arctan ( u ) - 1}{2 ( 1 + u ^{2} )}]_{- 1}^{1}

Compute the boundaries:

\frac{2 π + 4}{8}

= \frac{2 π + 4}{8}

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What is the integral from 0 to 2 of x/((x^2-2x+2)^2) ?
The integral from 0 to 2 of x/((x^2-2x+2)^2) is (2pi+4)/8