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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u + 1}{4 u ^{\frac{2}{3}}} d u

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

Expand

\frac{u + 1}{u ^{\frac{2}{3}}} : u^{\frac{1}{3}} + \frac{1}{u ^{\frac{2}{3}}}

= \frac{1}{4} \cdot \int u^{\frac{1}{3}} + \frac{1}{u ^{\frac{2}{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 u^{\frac{1}{3}} d u + \int \frac{1}{u ^{\frac{2}{3}}} d u)

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

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

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

Substitute back

u = 2 x - 1

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

Add a constant to the solution

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

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