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

1. Use integration by parts:

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

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

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

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

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

      $$\int 2 x\, dx = 2 \int x\, dx$$

      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: $x^{2}$

   Now evaluate the sub-integral.

2. Rewrite the integrand:

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

3. Integrate term-by-term:

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

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

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

      $$\int \left(-1\right)\, dx = - x$$

   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:

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

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

4. Now simplify:

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

5. Add the constant of integration:

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

The answer is:

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