Integral of (x^2)*e^(3x) dx

1. Use integration by parts:

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

   Let $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{3 x}$.

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

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

   1. There are multiple ways to do this integral.

      Method #1

      1. Let $u = 3 x$.

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

         $$\int \frac{e^{u}}{9}\, du$$

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

            $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

            1. The integral of the exponential function is itself.

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

            So, the result is: $\frac{e^{u}}{3}$

         Now substitute $u$ back in:

         $$\frac{e^{3 x}}{3}$$

      Method #2

      1. Let $u = e^{3 x}$.

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

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

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

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

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

               $$\int 1\, du = u$$

            So, the result is: $\frac{u}{3}$

         Now substitute $u$ back in:

         $$\frac{e^{3 x}}{3}$$

   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 $u{\left(x \right)} = \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = e^{3 x}$.

   Then $\operatorname{du}{\left(x \right)} = \frac{2}{3}$.

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

   1. Let $u = 3 x$.

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

      $$\int \frac{e^{u}}{9}\, du$$

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

         $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

         1. The integral of the exponential function is itself.

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

         So, the result is: $\frac{e^{u}}{3}$

      Now substitute $u$ back in:

      $$\frac{e^{3 x}}{3}$$

   Now evaluate the sub-integral.

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

   $$\int \frac{2 e^{3 x}}{9}\, dx = \frac{2 \int e^{3 x}\, dx}{9}$$

   1. Let $u = 3 x$.

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

      $$\int \frac{e^{u}}{9}\, du$$

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

         $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

         1. The integral of the exponential function is itself.

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

         So, the result is: $\frac{e^{u}}{3}$

      Now substitute $u$ back in:

      $$\frac{e^{3 x}}{3}$$

   So, the result is: $\frac{2 e^{3 x}}{27}$

4. Now simplify:

   $$\frac{\left(9 x^{2} - 6 x + 2\right) e^{3 x}}{27}$$

5. Add the constant of integration:

   $$\frac{\left(9 x^{2} - 6 x + 2\right) e^{3 x}}{27}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(9 x^{2} - 6 x + 2\right) e^{3 x}}{27}+ \mathrm{constant}$$