Integral of x/(cbrt(3x+1)) dx

1. Let $u = \sqrt[3]{3 x + 1}$.

   Then let $du = \frac{dx}{\left(3 x + 1\right)^{\frac{2}{3}}}$ and substitute $du$:

   $$\int u \left(\frac{u^{3}}{3} - \frac{1}{3}\right)\, du$$

   1. Rewrite the integrand:

      $$u \left(\frac{u^{3}}{3} - \frac{1}{3}\right) = \frac{u^{4}}{3} - \frac{u}{3}$$

   2. Integrate term-by-term:

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{u^{4}}{3}\, du = \frac{\int u^{4}\, du}{3}$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int u^{4}\, du = \frac{u^{5}}{5}$$

         So, the result is: $\frac{u^{5}}{15}$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \left(- \frac{u}{3}\right)\, du = - \frac{\int u\, du}{3}$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int u\, du = \frac{u^{2}}{2}$$

         So, the result is: $- \frac{u^{2}}{6}$

      The result is: $\frac{u^{5}}{15} - \frac{u^{2}}{6}$

   Now substitute $u$ back in:

   $$\frac{\left(3 x + 1\right)^{\frac{5}{3}}}{15} - \frac{\left(3 x + 1\right)^{\frac{2}{3}}}{6}$$

2. Now simplify:

   $$\frac{\left(2 x - 1\right) \left(3 x + 1\right)^{\frac{2}{3}}}{10}$$

3. Add the constant of integration:

   $$\frac{\left(2 x - 1\right) \left(3 x + 1\right)^{\frac{2}{3}}}{10}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(2 x - 1\right) \left(3 x + 1\right)^{\frac{2}{3}}}{10}+ \mathrm{constant}$$