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

The integral of x^2+6sqrt(x) is (x^3)/3+4x^{3/2}+C

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integral of x^2+6sqrt(x)

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

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

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

\int e^{t} - 3 d t

\int e^{- \frac{x}{6}} d x

\int e^{5 x - 1} d x

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