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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Add a constant to the solution

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

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