What is the integral of x^5(x^2+1)^2 ?

The integral of x^5(x^2+1)^2 is (x^{10}}{10}+\frac{x^8)/4+(x^6)/6+C

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integral of x^5(x^2+1)^2

\int x^{5} (x^{2} + 1)^{2} d x

\frac{x ^{10}}{10} + \frac{x ^{8}}{4} + \frac{x ^{6}}{6} + C

Solution steps

\int x^{5} (x^{2} + 1)^{2} d x

Expand

x^{5} (x^{2} + 1)^{2} : x^{9} + 2 x^{7} + x^{5}

= \int x^{9} + 2 x^{7} + 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 x^{9} d x + \int 2 x^{7} d x + \int x^{5} d x

\int x^{9} d x = \frac{x ^{10}}{10}

\int 2 x^{7} d x = \frac{x ^{8}}{4}

\int x^{5} d x = \frac{x ^{6}}{6}

= \frac{x ^{10}}{10} + \frac{x ^{8}}{4} + \frac{x ^{6}}{6}

Add a constant to the solution

= \frac{x ^{10}}{10} + \frac{x ^{8}}{4} + \frac{x ^{6}}{6} + C

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What is the integral of x^5(x^2+1)^2 ?
The integral of x^5(x^2+1)^2 is (x^{10}}{10}+\frac{x^8)/4+(x^6)/6+C