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

24e2x1+e2xdx

Short Answer

Expert verified

The solution of the given integral is 24e2x1+e2xdx=lne8-lne4.

Step by step solution

01

Step 1. Given Information 

Solving the given integrals.

24e2x1+e2xdx

02

Step 2. Using the substitution method. 

Let

u=1+e2xdudx=e2xdu=e2xdx

03

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

24e2x1+e2xdx=x=2x=41udu24e2x1+e2xdx=lnux=2x=424e2x1+e2xdx=lne2xx=2x=424e2x1+e2xdx=lne2×4-lne2×224e2x1+e2xdx=lne8-lne4

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