What is the integral from 0 to 4 of (x^2-1)/(x^2+2) ?

The integral from 0 to 4 of (x^2-1)/(x^2+2) is -3/(sqrt(2))arctan(2sqrt(2))+4

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integral from 0 to 4 of (x^2-1)/(x^2+2)

\int_{0}^{4} \frac{x ^{2} - 1}{x ^{2} + 2} d x

- \frac{3}{2} arctan (22) + 4

Decimal Notation

1.38874 \dots

Solution steps

\int_{0}^{4} \frac{x ^{2} - 1}{x ^{2} + 2} d x

\frac{x ^{2} - 1}{x ^{2} + 2} = - \frac{3}{x ^{2} + 2} + 1

= \int_{0}^{4} - \frac{3}{x ^{2} + 2} + 1 d x

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}^{4} \frac{3}{x ^{2} + 2} d x + \int_{0}^{4} 1 d x

\int_{0}^{4} \frac{3}{x ^{2} + 2} d x = \frac{3}{2} arctan (22)

\int_{0}^{4} 1 d x = 4

= - \frac{3}{2} arctan (22) + 4

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What is the integral from 0 to 4 of (x^2-1)/(x^2+2) ?
The integral from 0 to 4 of (x^2-1)/(x^2+2) is -3/(sqrt(2))arctan(2sqrt(2))+4