What is the integral of 1/(5\sqrt[5]{x^2)} ?

The integral of 1/(5\sqrt[5]{x^2)} is 1/3 x^{3/5}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(5\sqrt[5]{x^2)}

\frac{1}{3} x^{\frac{3}{5}} + 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 \frac{1}{x ^{\frac{2}{5}}} d x

Apply the Power Rule

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

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{3} x^{\frac{3}{5}} + C

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

\int sec^{- 2} (x) tan^{3} (x) d x

\int \frac{1}{49 - 25 x ^{2}} d x

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

\int \frac{ln ( x ^{2} )}{x} d x

What is the integral of 1/(5\sqrt[5]{x^2)} ?
The integral of 1/(5\sqrt[5]{x^2)} is 1/3 x^{3/5}+C