What is the integral of x/((x+4)^{1/3)} ?

The integral of x/((x+4)^{1/3)} is 3/5 (x+4)^{5/3}-6(x+4)^{2/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/((x+4)^{1/3)}

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

\frac{3}{5} (x + 4)^{\frac{5}{3}} - 6 (x + 4)^{\frac{2}{3}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 4}{u ^{\frac{1}{3}}} d u

= \int \frac{u - 4}{u ^{\frac{1}{3}}} d u

Expand

\frac{u - 4}{u ^{\frac{1}{3}}} : u^{\frac{2}{3}} - \frac{4}{u ^{\frac{1}{3}}}

= \int u^{\frac{2}{3}} - \frac{4}{u ^{\frac{1}{3}}} d u

Apply the Sum Rule:

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

= \int u^{\frac{2}{3}} d u - \int \frac{4}{u ^{\frac{1}{3}}} d u

\int u^{\frac{2}{3}} d u = \frac{3}{5} u^{\frac{5}{3}}

\int \frac{4}{u ^{\frac{1}{3}}} d u = 6 u^{\frac{2}{3}}

= \frac{3}{5} u^{\frac{5}{3}} - 6 u^{\frac{2}{3}}

Substitute back

u = x + 4

= \frac{3}{5} (x + 4)^{\frac{5}{3}} - 6 (x + 4)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{5} (x + 4)^{\frac{5}{3}} - 6 (x + 4)^{\frac{2}{3}} + C

f (x) = e^{x} + 1

\frac{8 t}{t ^{2} + 9}

\frac{( x - 1 )}{( x + 1 )}

\frac{d}{d x} (e^{0.5 x^{2}})

\frac{\partial}{\partial y} (x^{- 3} y^{2} + x y^{2} + 3 x y)

What is the integral of x/((x+4)^{1/3)} ?
The integral of x/((x+4)^{1/3)} is 3/5 (x+4)^{5/3}-6(x+4)^{2/3}+C