022 Exam 2 Sample B, Problem 5

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

Foundations:
This problem requires two rules of integration. In particular, you need
Integration by substitution (u - sub): If $u=g(x)$  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:

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

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

Step 3:
Now we need to substitute back into our original variables using our original substitution $u=e^{2x}+1$
to find  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \log(u) = \log(e^{2x} + 1).}

Step 4:
Since this integral is an indefinite integral we have to remember to add a constant  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} at the end.

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{2e^{2x}}{e^2x + 1}\, dx \,=\, \log(e^{2x}+1) + C.}

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