# Difference between revisions of "022 Exam 2 Sample A, Problem 3"

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

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