What is the integral of x^2+9sqrt(x) ?

The integral of x^2+9sqrt(x) is (x^3)/3+6x^{3/2}+C

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integral of x^2+9sqrt(x)

\int x^{2} + 9 x d x

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

Solution steps

\int x^{2} + 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^{2} d x + \int 9 x d x

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

\int 9 x d x = 6 x^{\frac{3}{2}}

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

Add a constant to the solution

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

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