What is the integral from 0 to a of (x^4+5)^3x^3 ?

The integral from 0 to a of (x^4+5)^3x^3 is (a^{16}}{16}+\frac{5a^{12})/4+(75a^8)/8+(125a^4)/4

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integral from 0 to a of (x^4+5)^3x^3

\int_{0}^{a} (x^{4} + 5)^{3} x^{3} d x

\frac{a ^{16}}{16} + \frac{5 a ^{12}}{4} + \frac{75 a ^{8}}{8} + \frac{125 a ^{4}}{4}

Solution steps

\int_{0}^{a} (x^{4} + 5)^{3} x^{3} d x

Expand

(x^{4} + 5)^{3} x^{3} : x^{15} + 15 x^{11} + 75 x^{7} + 125 x^{3}

= \int_{0}^{a} x^{15} + 15 x^{11} + 75 x^{7} + 125 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}^{a} x^{15} d x + \int_{0}^{a} 15 x^{11} d x + \int_{0}^{a} 75 x^{7} d x + \int_{0}^{a} 125 x^{3} d x

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

\int_{0}^{a} 15 x^{11} d x = \frac{5 a ^{12}}{4}

\int_{0}^{a} 75 x^{7} d x = \frac{75 a ^{8}}{8}

\int_{0}^{a} 125 x^{3} d x = \frac{125 a ^{4}}{4}

= \frac{a ^{16}}{16} + \frac{5 a ^{12}}{4} + \frac{75 a ^{8}}{8} + \frac{125 a ^{4}}{4}

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What is the integral from 0 to a of (x^4+5)^3x^3 ?
The integral from 0 to a of (x^4+5)^3x^3 is (a^{16}}{16}+\frac{5a^{12})/4+(75a^8)/8+(125a^4)/4