What is the integral from 0 to 2 of (x^6)/(1+x^2) ?

The integral from 0 to 2 of (x^6)/(1+x^2) is -arctan(2)+86/15

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integral from 0 to 2 of (x^6)/(1+x^2)

\int_{0}^{2} \frac{x ^{6}}{1 + x ^{2}} d x

- arctan (2) + \frac{86}{15}

Decimal Notation

4.62618 \dots

Solution steps

\int_{0}^{2} \frac{x ^{6}}{1 + x ^{2}} d x

Long division

\frac{x ^{6}}{1 + x ^{2}} : x^{4} - x^{2} + 1 - \frac{1}{x ^{2} + 1}

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

\int_{0}^{2} x^{4} d x = \frac{32}{5}

\int_{0}^{2} x^{2} d x = \frac{8}{3}

\int_{0}^{2} 1 d x = 2

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

= \frac{32}{5} - \frac{8}{3} + 2 - arctan (2)

Combine the fractions

2 + \frac{32}{5} - \frac{8}{3} : \frac{86}{15}

= - arctan (2) + \frac{86}{15}

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What is the integral from 0 to 2 of (x^6)/(1+x^2) ?
The integral from 0 to 2 of (x^6)/(1+x^2) is -arctan(2)+86/15