Integral of xe^(-4x) dx

1. Use integration by parts:

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

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

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

   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:

            $$\int \left(- \frac{e^{u}}{4}\right)\, du = - \frac{\int e^{u}\, du}{4}$$

            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:

            $$\int \left(- \frac{1}{4}\right)\, du = - \frac{\int 1\, du}{4}$$

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

   $$\int \left(- \frac{e^{- 4 x}}{4}\right)\, dx = - \frac{\int e^{- 4 x}\, dx}{4}$$

   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:

         $$\int \left(- \frac{e^{u}}{4}\right)\, du = - \frac{\int e^{u}\, du}{4}$$

         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: $\frac{e^{- 4 x}}{16}$

3. Now simplify:

   $$\frac{\left(- 4 x - 1\right) e^{- 4 x}}{16}$$

4. Add the constant of integration:

   $$\frac{\left(- 4 x - 1\right) e^{- 4 x}}{16}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(- 4 x - 1\right) e^{- 4 x}}{16}+ \mathrm{constant}$$