What is the integral from 0 to 1 of (x+1)/(x^2-2x+2) ?

The integral from 0 to 1 of (x+1)/(x^2-2x+2) is (-ln(2)+pi)/2

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integral from 0 to 1 of (x+1)/(x^2-2x+2)

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

\frac{- ln ( 2 ) + π}{2}

Decimal

1.22422 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

= - \frac{1}{2} ln (2) + \frac{π}{2}

Simplify

- \frac{1}{2} ln (2) + \frac{π}{2} : \frac{- ln ( 2 ) + π}{2}

= \frac{- ln ( 2 ) + π}{2}

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What is the integral from 0 to 1 of (x+1)/(x^2-2x+2) ?
The integral from 0 to 1 of (x+1)/(x^2-2x+2) is (-ln(2)+pi)/2