Finding antiderivitive of $\frac{x}{\sqrt[3]{1-3x}}$

Question

Evalute: $\int \frac{x}{\sqrt[3]{1-3x}}dx$

my try:

$u=1-3x$, $x=\frac{1-u}{3}, dx = - \frac{1}{3}du$

$\int \frac{x}{\sqrt[3]{1-3x}}dx = \int \frac{1-u}{3} \frac{1}{\sqrt[3]{u}}(-\frac{1}{3})du = -\frac{1}{9}\int \frac{1-u}{\sqrt[3]{u}}du = -\frac{1}{9}\int u^{-\frac{1}{3}}du + \frac{1}{9}\int u^{\frac{2}{3}}du=-\frac{1}{6} u^{\frac{2}{3}} + \frac{1}{15} u^{\frac{5}{3}} +C=-\frac{1}{6}(1-3x)^{\frac{2}{3}}+\frac{1}{15}(1-3x)^{\frac{5}{3}}+C$

I know I have done something wrong here, but cant figure out what, ps I'm not familiar with dx and du substitution

Answer 1

Your work seems correct.

Note that the final result can be written as: $$ -\frac{1}{6}(1-3x)^{\frac{2}{3}}+\frac{1}{15}(1-3x)^{\frac{5}{3}}+C= $$ $$ =-\frac{1}{6} \sqrt[3]{(1-3x)^2}+\frac{1}{15} \sqrt[3]{(1-3x)^5} +C= $$ $$ =\frac{1}{3}\sqrt[3]{(1-3x)^2}\left(-\frac{1}{2}+\frac{1}{5}-\frac{3}{5}x \right)+C= $$ $$ =-\frac{1}{5}\sqrt[3]{(1-3x)^2}\left(x+\frac{1}{2} \right)+C $$

Answer 2

$$y=1-3 x$$

$$x=\frac{1-y}{3}$$

$$\text{dx}=-\frac{\text{dy}}{3}$$

$$ x\text{dx}=\frac{1}{9} (y-1)\text{dy} $$

$$\frac{x}{\sqrt[3]{1-3 x}}\text{dx}=\frac{1}{9} (y-1) y^{-\frac{1}{3}}\text{dy}$$

$$\frac{1}{9} (y-1) y^{-\frac{1}{3}}\text{dy}=\frac{1}{9} \left(y^{2/3}-y^{-\frac{1}{3}}\right)\text{dy}$$

$$\frac{1}{9} \int \left(y^{2/3}-y^{-\frac{1}{3}}\right) \, dy=\frac{1}{3} \left(\frac{y}{5}-\frac{1}{2}\right) y^{2/3}$$

$$\frac{1}{3} \left(\frac{y}{5}-\frac{1}{2}\right) y^{2/3}=-\left(\frac{x}{5}+\frac{1}{10}\right) (1-3 x)^{2/3}=-\frac{1}{10} (2 x+1) \sqrt[3]{(1-3 x)^2}$$

$$\int \frac{x}{\sqrt[3]{1-3 x}} \, dx=-\frac{1}{10} (2 x+1) \sqrt[3]{(1-3 x)^2}+C$$