What is the integral from 0 to 1 of x^3(4+x^4)^3 ?

The integral from 0 to 1 of x^3(4+x^4)^3 is 369/16

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integral from 0 to 1 of x^3(4+x^4)^3

\int_{0}^{1} x^{3} (4 + x^{4})^{3} d x

\frac{369}{16}

Solution steps

\int_{0}^{1} x^{3} (4 + x^{4})^{3} d x

Expand

x^{3} (4 + x^{4})^{3} : 64 x^{3} + 48 x^{7} + 12 x^{11} + x^{15}

= \int_{0}^{1} 64 x^{3} + 48 x^{7} + 12 x^{11} + x^{15} 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} 64 x^{3} d x + \int_{0}^{1} 48 x^{7} d x + \int_{0}^{1} 12 x^{11} d x + \int_{0}^{1} x^{15} d x

\int_{0}^{1} 64 x^{3} d x = 16

\int_{0}^{1} 48 x^{7} d x = 6

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

\int_{0}^{1} x^{15} d x = \frac{1}{16}

= 16 + 6 + 1 + \frac{1}{16}

Simplify

16 + 6 + 1 + \frac{1}{16} : \frac{369}{16}

= \frac{369}{16}

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What is the integral from 0 to 1 of x^3(4+x^4)^3 ?
The integral from 0 to 1 of x^3(4+x^4)^3 is 369/16