Integral of x^2*exp(2*x) dx

1. Use integration by parts:

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

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

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

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

   1. Let $u = 2 x$.

      Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

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

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

         $$\text{False}$$

         1. The integral of the exponential function is itself.

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

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

      Now substitute $u$ back in:

      $$\frac{e^{2 x}}{2}$$

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

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

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

   1. Let $u = 2 x$.

      Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

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

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

         $$\text{False}$$

         1. The integral of the exponential function is itself.

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

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

      Now substitute $u$ back in:

      $$\frac{e^{2 x}}{2}$$

   Now evaluate the sub-integral.

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

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

   1. Let $u = 2 x$.

      Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

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

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

         $$\text{False}$$

         1. The integral of the exponential function is itself.

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

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

      Now substitute $u$ back in:

      $$\frac{e^{2 x}}{2}$$

   So, the result is: $\frac{e^{2 x}}{4}$

4. Now simplify:

   $$\frac{\left(2 x^{2} - 2 x + 1\right) e^{2 x}}{4}$$

5. Add the constant of integration:

   $$\frac{\left(2 x^{2} - 2 x + 1\right) e^{2 x}}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(2 x^{2} - 2 x + 1\right) e^{2 x}}{4}+ \mathrm{constant}$$