What is the integral of x(x^2+3)^2 ?

The integral of x(x^2+3)^2 is (x^6)/6+(3x^4)/2+(9x^2)/2+C

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integral of x(x^2+3)^2

\int x (x^{2} + 3)^{2} d x

\frac{x ^{6}}{6} + \frac{3 x ^{4}}{2} + \frac{9 x ^{2}}{2} + C

Solution steps

\int x (x^{2} + 3)^{2} d x

Expand

x (x^{2} + 3)^{2} : x^{5} + 6 x^{3} + 9 x

= \int x^{5} + 6 x^{3} + 9 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 6 x^{3} d x + \int 9 x d x

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

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

\int 9 x d x = \frac{9 x ^{2}}{2}

= \frac{x ^{6}}{6} + \frac{3 x ^{4}}{2} + \frac{9 x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{6}}{6} + \frac{3 x ^{4}}{2} + \frac{9 x ^{2}}{2} + C

\int \frac{2}{x ^{\frac{1}{2}}} d x

2 x + 3 y + (6 x + 9 y - 3) y^{^{'}} = 0

f (x) = cos (x^{2}) - 2 x^{2} sin (x^{2})

\frac{\partial}{\partial y} ((x + y) (x - y))

\frac{d}{d x} (tan^{40} (sec (cos (x))))

What is the integral of x(x^2+3)^2 ?
The integral of x(x^2+3)^2 is (x^6)/6+(3x^4)/2+(9x^2)/2+C