What is the integral of (z^4)/(\sqrt[3]{6+z^5)} ?

The integral of (z^4)/(\sqrt[3]{6+z^5)} is 3/10 (6+z^5)^{2/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (z^4)/(\sqrt[3]{6+z^5)}

\int \frac{z ^{4}}{3 6 + z ^{5}} d z

\frac{3}{10} (6 + z^{5})^{\frac{2}{3}} + C

Solution steps

\int \frac{z ^{4}}{3 6 + z ^{5}} d z

Apply u-substitution

: \int \frac{1}{5 3 u} d u

= \int \frac{1}{5 3 u} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{3 u} d u

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 = 6 + z^{5}

= \frac{1}{5} \cdot \frac{3}{2} (6 + z^{5})^{\frac{2}{3}}

Simplify

\frac{1}{5} \cdot \frac{3}{2} (6 + z^{5})^{\frac{2}{3}} : \frac{3}{10} (6 + z^{5})^{\frac{2}{3}}

= \frac{3}{10} (6 + z^{5})^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{10} (6 + z^{5})^{\frac{2}{3}} + C

\int sin^{5} (3 x) \cdot cos^{2} (3 x) d x

\int (4 - 7 m)^{\frac{1}{5}} d m

\int \frac{2}{9 - ( x + 3 ) ^{2}} d x

\int \frac{1}{2 y ^{3}} d y

\int x (e^{x} + 1) d x

What is the integral of (z^4)/(\sqrt[3]{6+z^5)} ?
The integral of (z^4)/(\sqrt[3]{6+z^5)} is 3/10 (6+z^5)^{2/3}+C