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

The integral of x/((4x+1)^{2/3)} is 1/16 (3/4 (4x+1)^{4/3}-3(4x+1)^{1/3})+C

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

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

\frac{1}{16} (\frac{3}{4} (4 x + 1)^{\frac{4}{3}} - 3 (4 x + 1)^{\frac{1}{3}}) + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{16 u ^{\frac{2}{3}}} d u

= \int \frac{u - 1}{16 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}{16} \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}{16} \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}{16} (\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}{16} (\frac{3}{4} u^{\frac{4}{3}} - 3 u^{\frac{1}{3}})

Substitute back

u = 4 x + 1

= \frac{1}{16} (\frac{3}{4} (4 x + 1)^{\frac{4}{3}} - 3 (4 x + 1)^{\frac{1}{3}})

Add a constant to the solution

= \frac{1}{16} (\frac{3}{4} (4 x + 1)^{\frac{4}{3}} - 3 (4 x + 1)^{\frac{1}{3}}) + C

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What is the integral of x/((4x+1)^{2/3)} ?
The integral of x/((4x+1)^{2/3)} is 1/16 (3/4 (4x+1)^{4/3}-3(4x+1)^{1/3})+C