Integral of (2x+1)*e^(-2x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 x$.

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

      $$\int \frac{\left(u + 1\right) e^{- u}}{2}\, du$$

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

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

         1. Let $u = - u$.

            Then let $du = - du$ and substitute $du$:

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

            1. Integrate term-by-term:

               1. Use integration by parts:

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

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

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

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

                  1. The integral of the exponential function is itself.

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

                  Now evaluate the sub-integral.

               2. The integral of the exponential function is itself.

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

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

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

                  1. The integral of the exponential function is itself.

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

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

               The result is: $u e^{u} - 2 e^{u}$

            Now substitute $u$ back in:

            $$- u e^{- u} - 2 e^{- u}$$

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

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

         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. Let $u = - 2 x$.

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

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

               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. 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 $- \frac{du}{2}$:

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

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

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

      1. Let $u = - 2 x$.

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

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

         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:

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

      The result is: $- x e^{- 2 x} - e^{- 2 x}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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