What is the integral from 0 to 1 of (x^2+1)^5x^3 ?

The integral from 0 to 1 of (x^2+1)^5x^3 is 107/28

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integral from 0 to 1 of (x^2+1)^5x^3

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

\frac{107}{28}

Decimal Notation

3.82142 \dots

Solution steps

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

Expand

(x^{2} + 1)^{5} x^{3} : x^{13} + 5 x^{11} + 10 x^{9} + 10 x^{7} + 5 x^{5} + x^{3}

= \int_{0}^{1} x^{13} + 5 x^{11} + 10 x^{9} + 10 x^{7} + 5 x^{5} + x^{3} 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^{13} d x + \int_{0}^{1} 5 x^{11} d x + \int_{0}^{1} 10 x^{9} d x + \int_{0}^{1} 10 x^{7} d x + \int_{0}^{1} 5 x^{5} d x + \int_{0}^{1} x^{3} d x

\int_{0}^{1} x^{13} d x = \frac{1}{14}

\int_{0}^{1} 5 x^{11} d x = \frac{5}{12}

\int_{0}^{1} 10 x^{9} d x = 1

\int_{0}^{1} 10 x^{7} d x = \frac{5}{4}

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

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

= \frac{1}{14} + \frac{5}{12} + 1 + \frac{5}{4} + \frac{5}{6} + \frac{1}{4}

Simplify

\frac{1}{14} + \frac{5}{12} + 1 + \frac{5}{4} + \frac{5}{6} + \frac{1}{4} : \frac{107}{28}

= \frac{107}{28}

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What is the integral from 0 to 1 of (x^2+1)^5x^3 ?
The integral from 0 to 1 of (x^2+1)^5x^3 is 107/28