022 Exam 2 Sample A, Problem 3

Find the antiderivative of $\int {\frac {1}{3x+2}}\,dx.$

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 $u=3x+2.$ This means $du=3dx$, and after substitution we have $\int {\frac {1}{3x+2}}=\int {\frac {1}{3u}}du$

Step 2:
We can now take the integral remembering the special rule:
$\int {\frac {1}{3u}}du={\frac {\log(u)}{3}}$

Step 3:
Now we need to substitute back into our original variables using our original substitution $u=3x+2$
to get ${\frac {\log(u)}{3}}={\frac {\log(3x+2)}{3}}$

Step 4:
Since this integral is an indefinite integral we have to remember to add "+ C" at the end.

Final Answer:
$\int {\frac {1}{3x+2}}dx={\frac {\ln(3x+2)}{3}}+C$

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