Question

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

${\int }_2^4\frac{e^{2x}}{1+e^{2x}}dx$

Short answer

The solution of the given integral is ${\int }_2^4\frac{e^{2x}}{1+e^{2x}}dx=\left[\ln e^8-\ln e^4\right]$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

${\int }_2^4\frac{e^{2x}}{1+e^{2x}}dx$

Step 2. Using the substitution method. 

Let

$$\begin{gathered} u=1+e^{2x} \\ \frac{du}{dx}=e^{2x} \\ du=e^{2x}dx \end{gathered}$$

Step 3. Using the information in equations, we can change variables completely:

$$\begin{gathered} {\int }_2^4\frac{e^{2x}}{1+e^{2x}}dx={\int }_{x=2}^{x=4}\frac{1}{u}du \\ {\int }_2^4\frac{e^{2x}}{1+e^{2x}}dx={\left[\ln u\right]}_{x=2}^{x=4} \\ {\int }_2^4\frac{e^{2x}}{1+e^{2x}}dx={\left[\ln e^{2x}\right]}_{x=2}^{x=4} \\ {\int }_2^4\frac{e^{2x}}{1+e^{2x}}dx=\left[\ln e^{2\times 4}-\ln e^{2\times 2}\right] \\ {\int }_2^4\frac{e^{2x}}{1+e^{2x}}dx=\left[\ln e^8-\ln e^4\right] \end{gathered}$$