What is the integral of 1/(\sqrt[5]{x^4)} ?

The integral of 1/(\sqrt[5]{x^4)} is 5\sqrt[5]{x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(\sqrt[5]{x^4)}

Solution steps

assuming

x \geq 0

= \int \frac{1}{x ^{\frac{4}{5}}} d x

Apply the Power Rule

: 5 x^{\frac{1}{5}}

= 5 x^{\frac{1}{5}}

Apply radical rule:

Add a constant to the solution

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

\int 2 (x^{2} + x + 2) e^{- 2 x} d x

cos^{3} (x)

\int (t er m o, i) d x

\int \frac{ln ^{4} ( x )}{x} d x

What is the integral of 1/(\sqrt[5]{x^4)} ?
The integral of 1/(\sqrt[5]{x^4)} is 5\sqrt[5]{x}+C