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

1. Use integration by parts:

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

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

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

   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:

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

   1. Rewrite the integrand:

      $$\frac{x^{3}}{x + 1} = x^{2} - x + 1 - \frac{1}{x + 1}$$

   2. Integrate term-by-term:

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

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

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

         $$\int \left(- x\right)\, dx = - \int x\, dx$$

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

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

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

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

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

         1. Let $u = x + 1$.

            Then let $du = dx$ and substitute $du$:

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

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

            Now substitute $u$ back in:

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

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

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

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

3. Add the constant of integration:

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

The answer is:

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