What is the integral of x(x^4+1)^2 ?

The integral of x(x^4+1)^2 is 1/2 x^{10}+x^6+1/2 x^2-(2x^{10})/5-(2x^6)/3+C

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integral of x(x^4+1)^2

\int x (x^{4} + 1)^{2} d x

\frac{1}{2} x^{10} + x^{6} + \frac{1}{2} x^{2} - \frac{2 x ^{10}}{5} - \frac{2 x ^{6}}{3} + C

Solution steps

\int x (x^{4} + 1)^{2} d x

Apply Integration By Parts

: \frac{1}{2} x^{2} (x^{4} + 1)^{2} - \int 4 x^{5} (x^{4} + 1) d x

= \frac{1}{2} x^{2} (x^{4} + 1)^{2} - \int 4 x^{5} (x^{4} + 1) d x

\int 4 x^{5} (x^{4} + 1) d x = 4 (\frac{x ^{10}}{10} + \frac{x ^{6}}{6})

= \frac{1}{2} x^{2} (x^{4} + 1)^{2} - 4 (\frac{x ^{10}}{10} + \frac{x ^{6}}{6})

Simplify

\frac{1}{2} x^{2} (x^{4} + 1)^{2} - 4 (\frac{x ^{10}}{10} + \frac{x ^{6}}{6}) : \frac{1}{2} x^{10} + x^{6} + \frac{1}{2} x^{2} - \frac{2 x ^{10}}{5} - \frac{2 x ^{6}}{3}

= \frac{1}{2} x^{10} + x^{6} + \frac{1}{2} x^{2} - \frac{2 x ^{10}}{5} - \frac{2 x ^{6}}{3}

Add a constant to the solution

= \frac{1}{2} x^{10} + x^{6} + \frac{1}{2} x^{2} - \frac{2 x ^{10}}{5} - \frac{2 x ^{6}}{3} + C

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What is the integral of x(x^4+1)^2 ?
The integral of x(x^4+1)^2 is 1/2 x^{10}+x^6+1/2 x^2-(2x^{10})/5-(2x^6)/3+C