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

The integral of x(x^2+1)^2 is (x^6)/6+(x^4)/2+(x^2)/2+C

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

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

\frac{x ^{6}}{6} + \frac{x ^{4}}{2} + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

x (x^{2} + 1)^{2} : x^{5} + 2 x^{3} + x

= \int x^{5} + 2 x^{3} + x 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^{5} d x + \int 2 x^{3} d x + \int x d x

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

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int x d x = \frac{x ^{2}}{2}

= \frac{x ^{6}}{6} + \frac{x ^{4}}{2} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{6}}{6} + \frac{x ^{4}}{2} + \frac{x ^{2}}{2} + C

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What is the integral of x(x^2+1)^2 ?
The integral of x(x^2+1)^2 is (x^6)/6+(x^4)/2+(x^2)/2+C