What is the integral of x(6+sqrt(x))^2 ?

The integral of x(6+sqrt(x))^2 is 18x^2+24/5 x^{5/2}+(x^3)/3+C

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integral of x(6+sqrt(x))^2

\int x (6 + x)^{2} d x

18 x^{2} + \frac{24}{5} x^{\frac{5}{2}} + \frac{x ^{3}}{3} + C

Solution steps

\int x (6 + x)^{2} d x

Expand

x (6 + x)^{2} : 36 x + 12 x^{\frac{3}{2}} + x^{2}

= \int 36 x + 12 x^{\frac{3}{2}} + x^{2} 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 36 x d x + \int 12 x^{\frac{3}{2}} d x + \int x^{2} d x

\int 36 x d x = 18 x^{2}

\int 12 x^{\frac{3}{2}} d x = \frac{24}{5} x^{\frac{5}{2}}

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

= 18 x^{2} + \frac{24}{5} x^{\frac{5}{2}} + \frac{x ^{3}}{3}

Add a constant to the solution

= 18 x^{2} + \frac{24}{5} x^{\frac{5}{2}} + \frac{x ^{3}}{3} + C

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What is the integral of x(6+sqrt(x))^2 ?
The integral of x(6+sqrt(x))^2 is 18x^2+24/5 x^{5/2}+(x^3)/3+C