Integral of (Ln^2(3x+1))/(3x+1) dx

1. Let $u = 3 x + 1$.

   Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

   $$\int \frac{\log{\left(u \right)}^{2}}{3 u}\, du$$

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

      $$\int \frac{\log{\left(u \right)}^{2}}{u}\, du = \frac{\int \frac{\log{\left(u \right)}^{2}}{u}\, du}{3}$$

      1. There are multiple ways to do this integral.

         Method #1

         1. Let $u = \frac{1}{u}$.

            Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$:

            $$\int \left(- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u}\right)\, du$$

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

               $$\int \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u}\, du$$

               1. Let $u = \log{\left(\frac{1}{u} \right)}$.

                  Then let $du = - \frac{du}{u}$ and substitute $- du$:

                  $$\int \left(- u^{2}\right)\, du$$

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

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

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

                        $$\int u^{2}\, du = \frac{u^{3}}{3}$$

                     So, the result is: $- \frac{u^{3}}{3}$

                  Now substitute $u$ back in:

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

               So, the result is: $\frac{\log{\left(\frac{1}{u} \right)}^{3}}{3}$

            Now substitute $u$ back in:

            $$\frac{\log{\left(u \right)}^{3}}{3}$$

         Method #2

         1. Let $u = \log{\left(u \right)}$.

            Then let $du = \frac{du}{u}$ and substitute $du$:

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

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

               $$\int u^{2}\, du = \frac{u^{3}}{3}$$

            Now substitute $u$ back in:

            $$\frac{\log{\left(u \right)}^{3}}{3}$$

      So, the result is: $\frac{\log{\left(u \right)}^{3}}{9}$

   Now substitute $u$ back in:

   $$\frac{\log{\left(3 x + 1 \right)}^{3}}{9}$$

2. Now simplify:

   $$\frac{\log{\left(3 x + 1 \right)}^{3}}{9}$$

3. Add the constant of integration:

   $$\frac{\log{\left(3 x + 1 \right)}^{3}}{9}+ \mathrm{constant}$$

The answer is:

$$\frac{\log{\left(3 x + 1 \right)}^{3}}{9}+ \mathrm{constant}$$