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

The integral of x/((2x+1)^{4/3)} is (6x+9)/(8(2x+1)^{1/3)}+C

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integral of x/((2x+1)^{4/3)}

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

\frac{6 x + 9}{8 ( 2 x + 1 ) ^{\frac{1}{3}}} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

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

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

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

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

Substitute back

u = 2 x + 1

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

Simplify

\frac{1}{4} (\frac{3}{2} (2 x + 1)^{\frac{2}{3}} - (- \frac{3}{( 2 x + 1 ) ^{\frac{1}{3}}})) : \frac{6 x + 9}{8 ( 2 x + 1 ) ^{\frac{1}{3}}}

= \frac{6 x + 9}{8 ( 2 x + 1 ) ^{\frac{1}{3}}}

Add a constant to the solution

= \frac{6 x + 9}{8 ( 2 x + 1 ) ^{\frac{1}{3}}} + C

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What is the integral of x/((2x+1)^{4/3)} ?
The integral of x/((2x+1)^{4/3)} is (6x+9)/(8(2x+1)^{1/3)}+C