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

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

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

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

- \frac{2}{3} + \frac{π}{4}

Decimal Notation

0.11873 \dots

Solution steps

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

Long division

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

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

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

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

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

= \frac{1}{3} - 1 + \frac{π}{4}

Simplify

\frac{1}{3} - 1 + \frac{π}{4} : - \frac{2}{3} + \frac{π}{4}

= - \frac{2}{3} + \frac{π}{4}

\int_{0}^{5} x e^{x} d x

\int_{0}^{2} x^{3} - 3 x^{2} + 2 d x

\int_{0}^{\frac{π}{4}} cos^{5} (2 x) d x

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

\int_{1}^{5} 5 x d x

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