What is the integral of (z^4)/(\sqrt[3]{3+z^5)} ?

The integral of (z^4)/(\sqrt[3]{3+z^5)} is 3/10 (3+z^5)^{2/3}+C

Popular >

integral of (z^4)/(\sqrt[3]{3+z^5)}

\frac{3}{10} (3 + z^{5})^{\frac{2}{3}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

= \frac{1}{5} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 3 + z^{5}

= \frac{1}{5} \cdot \frac{3}{2} (3 + z^{5})^{\frac{2}{3}}

Simplify

\frac{1}{5} \cdot \frac{3}{2} (3 + z^{5})^{\frac{2}{3}} : \frac{3}{10} (3 + z^{5})^{\frac{2}{3}}

= \frac{3}{10} (3 + z^{5})^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{10} (3 + z^{5})^{\frac{2}{3}} + C

\int cos (x) cos (2 x) sin (4 x) d x

\int \frac{1}{6 x + 8} d x

\int 2 e^{x} + 5 d x

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

\int \frac{1}{( 2 x + 5 ) ^{11}} d x

What is the integral of (z^4)/(\sqrt[3]{3+z^5)} ?
The integral of (z^4)/(\sqrt[3]{3+z^5)} is 3/10 (3+z^5)^{2/3}+C