Integral of x^2*e^(4*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 $\operatorname{dv}{\left(x \right)} = e^{4 x}$.

   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 $u = 4 x$.

         Then let $du = 4 dx$ and substitute $\frac{du}{4}$:

         $$\int \frac{e^{u}}{16}\, du$$

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

            $$\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$$

            1. The integral of the exponential function is itself.

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

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

         Now substitute $u$ back in:

         $$\frac{e^{4 x}}{4}$$

      Method #2

      1. Let $u = e^{4 x}$.

         Then let $du = 4 e^{4 x} dx$ and substitute $\frac{du}{4}$:

         $$\int \frac{1}{16}\, du$$

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

            $$\int \frac{1}{4}\, du = \frac{\int 1\, du}{4}$$

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

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

            So, the result is: $\frac{u}{4}$

         Now substitute $u$ back in:

         $$\frac{e^{4 x}}{4}$$

   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 $u{\left(x \right)} = \frac{x}{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{4 x}$.

   Then $\operatorname{du}{\left(x \right)} = \frac{1}{2}$.

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

   1. Let $u = 4 x$.

      Then let $du = 4 dx$ and substitute $\frac{du}{4}$:

      $$\int \frac{e^{u}}{16}\, du$$

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

         $$\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$$

         1. The integral of the exponential function is itself.

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

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

      Now substitute $u$ back in:

      $$\frac{e^{4 x}}{4}$$

   Now evaluate the sub-integral.

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

   $$\int \frac{e^{4 x}}{8}\, dx = \frac{\int e^{4 x}\, dx}{8}$$

   1. Let $u = 4 x$.

      Then let $du = 4 dx$ and substitute $\frac{du}{4}$:

      $$\int \frac{e^{u}}{16}\, du$$

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

         $$\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$$

         1. The integral of the exponential function is itself.

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

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

      Now substitute $u$ back in:

      $$\frac{e^{4 x}}{4}$$

   So, the result is: $\frac{e^{4 x}}{32}$

4. Now simplify:

   $$\frac{\left(8 x^{2} - 4 x + 1\right) e^{4 x}}{32}$$

5. Add the constant of integration:

   $$\frac{\left(8 x^{2} - 4 x + 1\right) e^{4 x}}{32}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(8 x^{2} - 4 x + 1\right) e^{4 x}}{32}+ \mathrm{constant}$$