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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=e^{2x}+1.} This means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=2e^{2x}\,dx} . After substitution we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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:
$\int {\frac {1}{u}}\,du\,=\,\ln(u).$

Step 3:
Now we need to substitute back into our original variables using our original substitution Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=e^{2x}+1}
to find  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \ln(u)=\ln(e^{2x}+1).}

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

Final Answer:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int {\frac {2e^{2x}}{e^{2}x+1}}\,dx\,=\,\ln(e^{2x}+1)+C.}

Return to Sample Exam