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

The integral from 0 to 1 of x(x^2+1)^2 is 7/6

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

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

\frac{7}{6}

Decimal

1.16666 \dots

Solution steps

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

Expand

x (x^{2} + 1)^{2} : x^{5} + 2 x^{3} + x

= \int_{0}^{1} x^{5} + 2 x^{3} + x 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^{5} d x + \int_{0}^{1} 2 x^{3} d x + \int_{0}^{1} x d x

\int_{0}^{1} x^{5} d x = \frac{1}{6}

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

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

= \frac{1}{6} + \frac{1}{2} + \frac{1}{2}

Simplify

\frac{1}{6} + \frac{1}{2} + \frac{1}{2} : \frac{7}{6}

= \frac{7}{6}

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What is the integral from 0 to 1 of x(x^2+1)^2 ?
The integral from 0 to 1 of x(x^2+1)^2 is 7/6