What is the integral of 8y-2/(y^{1/4)} ?

The integral of 8y-2/(y^{1/4)} is 4y^2-8/3 y^{3/4}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 8y-2/(y^{1/4)}

\int 8 y - \frac{2}{y ^{\frac{1}{4}}} d y

4 y^{2} - \frac{8}{3} y^{\frac{3}{4}} + C

Solution steps

\int 8 y - \frac{2}{y ^{\frac{1}{4}}} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 8 y d y - \int \frac{2}{y ^{\frac{1}{4}}} d y

\int 8 y d y = 4 y^{2}

\int \frac{2}{y ^{\frac{1}{4}}} d y = \frac{8}{3} y^{\frac{3}{4}}

= 4 y^{2} - \frac{8}{3} y^{\frac{3}{4}}

Add a constant to the solution

= 4 y^{2} - \frac{8}{3} y^{\frac{3}{4}} + C

What is the integral of 8y-2/(y^{1/4)} ?
The integral of 8y-2/(y^{1/4)} is 4y^2-8/3 y^{3/4}+C