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

The integral from 0 to 2 of (x^2-x)/(x^2+2) is -sqrt(2)arctan(sqrt(2))+2-1/2 ln(3)

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

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

- 2 arctan (2) + 2 - \frac{1}{2} ln (3)

Decimal Notation

0.09967 \dots

Solution steps

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

Expand

\frac{x ^{2} - x}{x ^{2} + 2} : \frac{x ^{2}}{x ^{2} + 2} - \frac{x}{x ^{2} + 2}

Apply the Sum Rule:

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

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

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

\int_{0}^{2} \frac{x}{x ^{2} + 2} d x = \frac{1}{2} ln (3)

= - 2 arctan (2) + 2 - \frac{1}{2} ln (3)

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What is the integral from 0 to 2 of (x^2-x)/(x^2+2) ?
The integral from 0 to 2 of (x^2-x)/(x^2+2) is -sqrt(2)arctan(sqrt(2))+2-1/2 ln(3)