Integral of x^2ln(12x+1) dx

1. Use integration by parts:

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

   Let $u{\left(x \right)} = \log{\left(12 x + 1 \right)}$ and let $\operatorname{dv}{\left(x \right)} = x^{2}$.

   Then $\operatorname{du}{\left(x \right)} = \frac{12}{12 x + 1}$.

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

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

      $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

   Now evaluate the sub-integral.

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

   $$\int \frac{4 x^{3}}{12 x + 1}\, dx = 4 \int \frac{x^{3}}{12 x + 1}\, dx$$

   1. Rewrite the integrand:

      $$\frac{x^{3}}{12 x + 1} = \frac{x^{2}}{12} - \frac{x}{144} + \frac{1}{1728} - \frac{1}{1728 \left(12 x + 1\right)}$$

   2. Integrate term-by-term:

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

         $$\int \frac{x^{2}}{12}\, dx = \frac{\int x^{2}\, dx}{12}$$

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

            $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

         So, the result is: $\frac{x^{3}}{36}$

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

         $$\int \left(- \frac{x}{144}\right)\, dx = - \frac{\int x\, dx}{144}$$

         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: $- \frac{x^{2}}{288}$

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

         $$\int \frac{1}{1728}\, dx = \frac{x}{1728}$$

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

         $$\int \left(- \frac{1}{1728 \left(12 x + 1\right)}\right)\, dx = - \frac{\int \frac{1}{12 x + 1}\, dx}{1728}$$

         1. Let $u = 12 x + 1$.

            Then let $du = 12 dx$ and substitute $\frac{du}{12}$:

            $$\int \frac{1}{12 u}\, du$$

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

               $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{12}$$

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

               So, the result is: $\frac{\log{\left(u \right)}}{12}$

            Now substitute $u$ back in:

            $$\frac{\log{\left(12 x + 1 \right)}}{12}$$

         So, the result is: $- \frac{\log{\left(12 x + 1 \right)}}{20736}$

      The result is: $\frac{x^{3}}{36} - \frac{x^{2}}{288} + \frac{x}{1728} - \frac{\log{\left(12 x + 1 \right)}}{20736}$

   So, the result is: $\frac{x^{3}}{9} - \frac{x^{2}}{72} + \frac{x}{432} - \frac{\log{\left(12 x + 1 \right)}}{5184}$

3. Now simplify:

   $$\frac{x^{3} \log{\left(12 x + 1 \right)}}{3} - \frac{x^{3}}{9} + \frac{x^{2}}{72} - \frac{x}{432} + \frac{\log{\left(12 x + 1 \right)}}{5184}$$

4. Add the constant of integration:

   $$\frac{x^{3} \log{\left(12 x + 1 \right)}}{3} - \frac{x^{3}}{9} + \frac{x^{2}}{72} - \frac{x}{432} + \frac{\log{\left(12 x + 1 \right)}}{5184}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{3} \log{\left(12 x + 1 \right)}}{3} - \frac{x^{3}}{9} + \frac{x^{2}}{72} - \frac{x}{432} + \frac{\log{\left(12 x + 1 \right)}}{5184}+ \mathrm{constant}$$