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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 3/(\sqrt[5]{x^2)}

5 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

= 3 \cdot \int \frac{1}{x ^{\frac{2}{5}}} d x

Apply the Power Rule

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

Simplify

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

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

Add a constant to the solution

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

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