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

The integral of (z^3)/(\sqrt[3]{3+z^4)} is 3/8 (3+z^4)^{2/3}+C

Popular >

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

\frac{3}{8} (3 + z^{4})^{\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}{4} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 3 + z^{4}

= \frac{1}{4} \cdot \frac{3}{2} (3 + z^{4})^{\frac{2}{3}}

Simplify

\frac{1}{4} \cdot \frac{3}{2} (3 + z^{4})^{\frac{2}{3}} : \frac{3}{8} (3 + z^{4})^{\frac{2}{3}}

= \frac{3}{8} (3 + z^{4})^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{8} (3 + z^{4})^{\frac{2}{3}} + C

\int \frac{( 1 + x )}{1 - x} d x

\int x^{5} ln (2 x^{3}) d x

\int \frac{x ^{3} - 2 x ^{2} + x - 1}{x ^{2} - 1} d x

\int cos (m x) cos (n x) d x

What is the integral of (z^3)/(\sqrt[3]{3+z^4)} ?
The integral of (z^3)/(\sqrt[3]{3+z^4)} is 3/8 (3+z^4)^{2/3}+C