Integral of e^(-x)*(e^(2*x)+3*e^x) dx

1. There are multiple ways to do this integral.

   Method #1

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

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

      $$\int \frac{- 3 u - 1}{u^{2}}\, du$$

      1. Rewrite the integrand:

         $$\frac{- 3 u - 1}{u^{2}} = - \frac{3}{u} - \frac{1}{u^{2}}$$

      2. Integrate term-by-term:

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

            $$\int \left(- \frac{3}{u}\right)\, du = - 3 \int \frac{1}{u}\, du$$

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            So, the result is: $- 3 \log{\left(u \right)}$

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

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

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

            So, the result is: $\frac{1}{u}$

         The result is: $- 3 \log{\left(u \right)} + \frac{1}{u}$

      Now substitute $u$ back in:

      $$e^{x} - 3 \log{\left(e^{- x} \right)}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(3 e^{x} + e^{2 x}\right) e^{- x} = e^{x} + 3$$

   2. Integrate term-by-term:

      1. The integral of the exponential function is itself.

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

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

         $$\int 3\, dx = 3 x$$

      The result is: $3 x + e^{x}$

2. Now simplify:

   $$e^{x} - 3 \log{\left(e^{- x} \right)}$$

3. Add the constant of integration:

   $$e^{x} - 3 \log{\left(e^{- x} \right)}+ \mathrm{constant}$$

The answer is:

$$e^{x} - 3 \log{\left(e^{- x} \right)}+ \mathrm{constant}$$