What is the integral from 0 to 1 of (3+x^2)^2x ?

The integral from 0 to 1 of (3+x^2)^2x is 37/6

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integral from 0 to 1 of (3+x^2)^2x

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

\frac{37}{6}

Decimal

6.16666 \dots

Solution steps

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

Expand

(3 + x^{2})^{2} x : 9 x + 6 x^{3} + x^{5}

= \int_{0}^{1} 9 x + 6 x^{3} + x^{5} 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} 9 x d x + \int_{0}^{1} 6 x^{3} d x + \int_{0}^{1} x^{5} d x

\int_{0}^{1} 9 x d x = \frac{9}{2}

\int_{0}^{1} 6 x^{3} d x = \frac{3}{2}

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

= \frac{9}{2} + \frac{3}{2} + \frac{1}{6}

Simplify

\frac{9}{2} + \frac{3}{2} + \frac{1}{6} : \frac{37}{6}

= \frac{37}{6}

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What is the integral from 0 to 1 of (3+x^2)^2x ?
The integral from 0 to 1 of (3+x^2)^2x is 37/6