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

The integral of x/((3x+1)^{1/3)} is 1/10 (2x-1)(3x+1)^{2/3}+C

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

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

\frac{1}{10} (2 x - 1) (3 x + 1)^{\frac{2}{3}} + C

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply u-substitution

: \int \frac{u - 1}{9 u ^{\frac{1}{3}}} d u

= \int \frac{u - 1}{9 u ^{\frac{1}{3}}} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{u - 1}{u ^{\frac{1}{3}}} d u

Expand

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

= \frac{1}{9} \cdot \int u^{\frac{2}{3}} - \frac{1}{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

= \frac{1}{9} (\int u^{\frac{2}{3}} d u - \int \frac{1}{u ^{\frac{1}{3}}} d u)

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

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

= \frac{1}{9} (\frac{3}{5} u^{\frac{5}{3}} - \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = 3 x + 1

= \frac{1}{9} (\frac{3}{5} (3 x + 1)^{\frac{5}{3}} - \frac{3}{2} (3 x + 1)^{\frac{2}{3}})

Simplify

\frac{1}{9} (\frac{3}{5} (3 x + 1)^{\frac{5}{3}} - \frac{3}{2} (3 x + 1)^{\frac{2}{3}}) : \frac{1}{10} (2 x - 1) (3 x + 1)^{\frac{2}{3}}

= \frac{1}{10} (2 x - 1) (3 x + 1)^{\frac{2}{3}}

Add a constant to the solution

= \frac{1}{10} (2 x - 1) (3 x + 1)^{\frac{2}{3}} + C

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What is the integral of x/((3x+1)^{1/3)} ?
The integral of x/((3x+1)^{1/3)} is 1/10 (2x-1)(3x+1)^{2/3}+C