Integral of 4*x*e^(-2x) dx

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

   $$\int 4 x e^{- 2 x}\, dx = 4 \int x e^{- 2 x}\, dx$$

   1. Use integration by parts:

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

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

      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 $u = - 2 x$.

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

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

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

               $$\int \left(- \frac{e^{u}}{2}\right)\, du = - \frac{\int e^{u}\, du}{2}$$

               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:

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

         Method #2

         1. Let $u = e^{- 2 x}$.

            Then let $du = - 2 e^{- 2 x} dx$ and substitute $- \frac{du}{2}$:

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

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

               $$\int \left(- \frac{1}{2}\right)\, du = - \frac{\int 1\, du}{2}$$

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

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

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

            Now substitute $u$ back in:

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

      Now evaluate the sub-integral.

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

      $$\int \left(- \frac{e^{- 2 x}}{2}\right)\, dx = - \frac{\int e^{- 2 x}\, dx}{2}$$

      1. Let $u = - 2 x$.

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

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

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

            $$\int \left(- \frac{e^{u}}{2}\right)\, du = - \frac{\int e^{u}\, du}{2}$$

            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: $\frac{e^{- 2 x}}{4}$

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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