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

The integral of 1/(x^{2/5)} is 5/3 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/(x^{2/5)}

\int \frac{1}{x ^{\frac{2}{5}}} d x

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

Solution steps

\int \frac{1}{x ^{\frac{2}{5}}} d x

Apply the Power Rule

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

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

Add a constant to the solution

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

\int_{0}^{2} x^{2} d x

\int e^{x^{2}} x^{2} d x

y^{^{'}} = e^{5 y + 8 x}, y (0) = 3

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

\frac{d}{d x} (\frac{2 x ^{4} + 3 x ^{2} - 1}{x ^{2}})

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