Integral of (log2(3x+1))/3x+1 dx

1. Integrate term-by-term:

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

      $$\int \frac{x \log{\left(3 x + 1 \right)}}{3 \log{\left(2 \right)}}\, dx = \frac{\int x \log{\left(3 x + 1 \right)}\, dx}{3 \log{\left(2 \right)}}$$

      1. Use integration by parts:

         $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

         Let $u{\left(x \right)} = \log{\left(3 x + 1 \right)}$ and let $\operatorname{dv}{\left(x \right)} = x$.

         Then $\operatorname{du}{\left(x \right)} = \frac{3}{3 x + 1}$.

         To find $v{\left(x \right)}$:

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

            $$\int x\, dx = \frac{x^{2}}{2}$$

         Now evaluate the sub-integral.

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

         $$\int \frac{3 x^{2}}{2 \cdot \left(3 x + 1\right)}\, dx = \frac{3 \int \frac{x^{2}}{3 x + 1}\, dx}{2}$$

         1. Rewrite the integrand:

            $$\frac{x^{2}}{3 x + 1} = \frac{x}{3} - \frac{1}{9} + \frac{1}{9 \cdot \left(3 x + 1\right)}$$

         2. Integrate term-by-term:

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

               $$\int \frac{x}{3}\, dx = \frac{\int x\, dx}{3}$$

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

                  $$\int x\, dx = \frac{x^{2}}{2}$$

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

            1. The integral of a constant is the constant times the variable of integration:

               $$\int \left(- \frac{1}{9}\right)\, dx = - \frac{x}{9}$$

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

               $$\int \frac{1}{9 \cdot \left(3 x + 1\right)}\, dx = \frac{\int \frac{1}{3 x + 1}\, dx}{9}$$

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

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

                  $$\int \frac{1}{9 u}\, du$$

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

                     $$\int \frac{1}{3 u}\, du = \frac{\int \frac{1}{u}\, du}{3}$$

                     1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

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

                  Now substitute $u$ back in:

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

               So, the result is: $\frac{\log{\left(3 x + 1 \right)}}{27}$

            The result is: $\frac{x^{2}}{6} - \frac{x}{9} + \frac{\log{\left(3 x + 1 \right)}}{27}$

         So, the result is: $\frac{x^{2}}{4} - \frac{x}{6} + \frac{\log{\left(3 x + 1 \right)}}{18}$

      So, the result is: $\frac{\frac{x^{2} \log{\left(3 x + 1 \right)}}{2} - \frac{x^{2}}{4} + \frac{x}{6} - \frac{\log{\left(3 x + 1 \right)}}{18}}{3 \log{\left(2 \right)}}$

   1. The integral of a constant is the constant times the variable of integration:

      $$\int 1\, dx = x$$

   The result is: $x + \frac{\frac{x^{2} \log{\left(3 x + 1 \right)}}{2} - \frac{x^{2}}{4} + \frac{x}{6} - \frac{\log{\left(3 x + 1 \right)}}{18}}{3 \log{\left(2 \right)}}$

2. Now simplify:

   $$\frac{18 x^{2} \log{\left(3 x + 1 \right)} - 9 x^{2} + 6 x + x \log{\left(324518553658426726783156020576256 \right)} - 2 \log{\left(3 x + 1 \right)}}{108 \log{\left(2 \right)}}$$

3. Add the constant of integration:

   $$\frac{18 x^{2} \log{\left(3 x + 1 \right)} - 9 x^{2} + 6 x + x \log{\left(324518553658426726783156020576256 \right)} - 2 \log{\left(3 x + 1 \right)}}{108 \log{\left(2 \right)}}+ \mathrm{constant}$$

The answer is: