What is the integral of \sqrt[7]{y^5} ?

The integral of \sqrt[7]{y^5} is 7/12 y^{12/7}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of \sqrt[7]{y^5}

\frac{7}{12} y^{\frac{12}{7}} + C

Solution steps

assuming

y \geq 0

= \int y^{\frac{5}{7}} d y

Apply the Power Rule

: \frac{7}{12} y^{\frac{12}{7}}

= \frac{7}{12} y^{\frac{12}{7}}

Add a constant to the solution

= \frac{7}{12} y^{\frac{12}{7}} + C

\int \frac{1}{- x ^{2} + 2 x} d x

\int_{0}^{0.4} e^{- x^{2}} d x

(3 + x^{2}) d y + 2 \cdot x \cdot y \cdot d x = cos (x) d x

\int \frac{1}{( 6 - 5 x ) ^{2}} d x

\int e^{- 2 t} + 2 d t

What is the integral of \sqrt[7]{y^5} ?
The integral of \sqrt[7]{y^5} is 7/12 y^{12/7}+C