What is the integral of \sqrt[3]{2x-4} ?

The integral of \sqrt[3]{2x-4} is 3/8 (2x-4)^{4/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of \sqrt[3]{2x-4}

\frac{3}{8} (2 x - 4)^{\frac{4}{3}} + C

Solution steps

Apply u-substitution

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= \frac{1}{2} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 2 x - 4

= \frac{1}{2} \cdot \frac{3}{4} (2 x - 4)^{\frac{4}{3}}

Simplify

\frac{1}{2} \cdot \frac{3}{4} (2 x - 4)^{\frac{4}{3}} : \frac{3}{8} (2 x - 4)^{\frac{4}{3}}

= \frac{3}{8} (2 x - 4)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{8} (2 x - 4)^{\frac{4}{3}} + C

\int \frac{e ^{7 x}}{e ^{7 x} + 2} d x

\int 6 sec^{2} (x) tan^{2} (x) d x

\int \frac{- 4 n - 2}{( n - 3 ) ^{2}} d n

\int cos^{2} (x) cos (x) d x

What is the integral of \sqrt[3]{2x-4} ?
The integral of \sqrt[3]{2x-4} is 3/8 (2x-4)^{4/3}+C