# 022 Exam 2 Sample B, Problem 5

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Find the antiderivative of ${\displaystyle \int {\frac {2e^{2x}}{e^{2x}+1}}\,dx.}$

Foundations:
This problem requires two rules of integration. In particular, you need
Integration by substitution (u - sub): If ${\displaystyle u=g(x)}$  is a differentiable functions whose range is in the domain of ${\displaystyle f}$, then
${\displaystyle \int g'(x)f(g(x))dx\,=\,\int f(u)du.}$
We also need the derivative of the natural log since we will recover natural log from integration:
${\displaystyle \left(ln(x)\right)'\,=\,{\frac {1}{x}}.}$

Solution:

Step 1:
Use a u-substitution with ${\displaystyle u=e^{2x}+1.}$ This means ${\displaystyle du=2e^{2x}\,dx}$. After substitution we have
${\displaystyle \int {\frac {2e^{2x}}{e^{2x}+1}}\,dx\,=\,\int {\frac {1}{u}}\,du.}$
Step 2:
We can now take the integral remembering the special rule:
${\displaystyle \int {\frac {1}{u}}\,du\,=\,\ln(u).}$
Step 3:
Now we need to substitute back into our original variables using our original substitution ${\displaystyle u=e^{2x}+1}$
to find  ${\displaystyle \ln(u)=\ln(e^{2x}+1).}$
Step 4:
Since this integral is an indefinite integral we have to remember to add a constant  ${\displaystyle C}$ at the end.
${\displaystyle \int {\frac {2e^{2x}}{e^{2x}+1}}\,dx\,=\,\ln(e^{2x}+1)+C.}$