What is the integral of x^{1/3}+9 ?

The integral of x^{1/3}+9 is 3/4 x^{4/3}+9x+C

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integral of x^{1/3}+9

\int x^{\frac{1}{3}} + 9 d x

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

Solution steps

\int x^{\frac{1}{3}} + 9 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^{\frac{1}{3}} d x + \int 9 d x

\int x^{\frac{1}{3}} d x = \frac{3}{4} x^{\frac{4}{3}}

\int 9 d x = 9 x

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

Add a constant to the solution

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

What is the integral of x^{1/3}+9 ?
The integral of x^{1/3}+9 is 3/4 x^{4/3}+9x+C