Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int \frac{e^x}{2-e^x}\mathrm{d}x$

Short answer

The solution of the given integral is $\int \frac{e^x}{2-e^x}\mathrm{d}x=-\ln \left(2-{\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{C}$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int \frac{e^x}{2-e^x}\mathrm{d}x$

Step 2. Solving the given integral using substitution method. 

Let

$$\begin{gathered} u=2-e^x \\ \frac{du}{dx}=-e^x \\ du=-e^{x}dx \\ -du=e^{x}dx \end{gathered}$$

Step 3. This substitution changes the integral into  

$$\begin{gathered} \int \frac{e^x}{2-e^x}\mathrm{d}x=-\int \frac{1}{u}\mathrm{du} \\ \int \frac{{\mathrm{e}}^{\mathrm{x}}}{2-{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}=-\ln \left(\mathrm{u}\right)+\mathrm{C} \\ \int \frac{{\mathrm{e}}^{\mathrm{x}}}{2-{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}=-\ln \left(2-{\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{C} \end{gathered}$$