Integral of ln^2(x+3) dx

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

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

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

   1. Use integration by parts:

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

      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

      Then $\operatorname{du}{\left(u \right)} = 2 u$.

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

      1. The integral of the exponential function is itself.

         $$\int e^{u}\, du = e^{u}$$

      Now evaluate the sub-integral.

   2. Use integration by parts:

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

      Let $u{\left(u \right)} = 2 u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

      Then $\operatorname{du}{\left(u \right)} = 2$.

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

      1. The integral of the exponential function is itself.

         $$\int e^{u}\, du = e^{u}$$

      Now evaluate the sub-integral.

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

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

      1. The integral of the exponential function is itself.

         $$\int e^{u}\, du = e^{u}$$

      So, the result is: $2 e^{u}$

   Now substitute $u$ back in:

   $$2 x + \left(x + 3\right) \log{\left(x + 3 \right)}^{2} - 2 \left(x + 3\right) \log{\left(x + 3 \right)} + 6$$

2. Now simplify:

   $$2 x + \left(x + 3\right) \log{\left(x + 3 \right)}^{2} - 2 \left(x + 3\right) \log{\left(x + 3 \right)} + 6$$

3. Add the constant of integration:

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

The answer is: