022 Exam 2 Sample B, Problem 5

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

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

 Solution:

ExpandStep 1:
Use a u-substitution with $u=e^{2x}+1.$ This means $du=2e^{2x}\,dx$. After substitution we have $\int {\frac {2e^{2x}}{e^{2x}+1}}\,dx\,=\,\int {\frac {1}{u}}\,du.$

ExpandStep 2:
We can now take the integral remembering the special rule:
$\int {\frac {1}{u}}\,du\,=\,\log(u).$

ExpandStep 3:
Now we need to substitute back into our original variables using our original substitution $u=e^{2x}+1$
to find  $\log(u)=\log(e^{2x}+1).$

ExpandStep 4:
Since this integral is an indefinite integral we have to remember to add a constant  $C$ at the end.

ExpandFinal Answer:
$\int {\frac {2e^{2x}}{e^{2}x+1}}\,dx\,=\,\log(e^{2x}+1)+C.$

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