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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x+1)^{-1/3}

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

= \frac{x}{( 2 x + 1 ) ^{\frac{1}{3}}} + \frac{2 x + 3}{4 ( 2 x + 1 ) ^{\frac{1}{3}}} + C

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

\int i n^{5} x d x

\int \frac{- 3 ln ( x )}{x} d x

\int \frac{t}{3} d t

\int \frac{2 x + 1}{x ( x - 2 )} d x

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