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

The integral of x(x^2+4)^2 is (x^6)/6+2x^4+8x^2+C

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

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

\frac{x ^{6}}{6} + 2 x^{4} + 8 x^{2} + C

Solution steps

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

Expand

x (x^{2} + 4)^{2} : x^{5} + 8 x^{3} + 16 x

= \int x^{5} + 8 x^{3} + 16 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 8 x^{3} d x + \int 16 x d x

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

\int 8 x^{3} d x = 2 x^{4}

\int 16 x d x = 8 x^{2}

= \frac{x ^{6}}{6} + 2 x^{4} + 8 x^{2}

Add a constant to the solution

= \frac{x ^{6}}{6} + 2 x^{4} + 8 x^{2} + C

\int \frac{sec ^{4} ( t )}{tan ^{2} ( t )} d t

\int \frac{tan ( ln ( 5 x + 6 ) )}{5 x + 6} d x

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

\frac{d}{d x} (\frac{y}{x ^{2} + y ^{2}})

n = 0 \sum \infty e^{2}

What is the integral of x(x^2+4)^2 ?
The integral of x(x^2+4)^2 is (x^6)/6+2x^4+8x^2+C