What is the integral of (9sqrt(x^3)+4\sqrt[3]{x^2}) ?

The integral of (9sqrt(x^3)+4\sqrt[3]{x^2}) is 18/5 x^{5/2}+12/5 x^{5/3}+C

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integral of (9sqrt(x^3)+4\sqrt[3]{x^2})

\frac{18}{5} x^{\frac{5}{2}} + \frac{12}{5} x^{\frac{5}{3}} + C

Solution steps

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int 9 x^{3} d x = \frac{18}{5} x^{\frac{5}{2}}

= \frac{18}{5} x^{\frac{5}{2}} + \frac{12}{5} x^{\frac{5}{3}}

Add a constant to the solution

= \frac{18}{5} x^{\frac{5}{2}} + \frac{12}{5} x^{\frac{5}{3}} + C

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What is the integral of (9sqrt(x^3)+4\sqrt[3]{x^2}) ?
The integral of (9sqrt(x^3)+4\sqrt[3]{x^2}) is 18/5 x^{5/2}+12/5 x^{5/3}+C