What is the integral of-(\sqrt[3]{x^2})/5 ?

The integral of-(\sqrt[3]{x^2})/5 is -3/25 x^{5/3}+C

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integral of-(\sqrt[3]{x^2})/5

- \frac{3}{25} x^{\frac{5}{3}} + C

Solution steps

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

assuming

x \geq 0

= - \frac{1}{5} \cdot \int x^{\frac{2}{3}} d x

Apply the Power Rule

: \frac{3}{5} x^{\frac{5}{3}}

= - \frac{1}{5} \cdot \frac{3}{5} x^{\frac{5}{3}}

Simplify

- \frac{1}{5} \cdot \frac{3}{5} x^{\frac{5}{3}} : - \frac{3}{25} x^{\frac{5}{3}}

= - \frac{3}{25} x^{\frac{5}{3}}

Add a constant to the solution

= - \frac{3}{25} x^{\frac{5}{3}} + C

\int \frac{4 + x}{4 - x} d x

\frac{\partial}{\partial y} (x^{4} - 3 x^{2} y)

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

\frac{8 x ^{5} - 15}{x ^{4}}

\frac{1}{( x ^{2} + 1 )}

What is the integral of-(\sqrt[3]{x^2})/5 ?
The integral of-(\sqrt[3]{x^2})/5 is -3/25 x^{5/3}+C