# 022 Exam 2 Sample A, Problem 3

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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 ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions, then

${\displaystyle (f\circ g)'(x)=f'(g(x))\cdot g'(x).}$

The Product Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions, then

${\displaystyle (fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).}$

The Quotient Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions and ${\displaystyle g(x)\neq 0}$ , then

${\displaystyle \left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.}$
Additionally, we will need our power rule for differentiation:
${\displaystyle \left(x^{n}\right)'\,=\,nx^{n-1},}$ for ${\displaystyle n\neq 0}$,
as well as the derivative of natural log:
${\displaystyle \left(\ln x\right)'\,=\,{\frac {1}{x}}.}$

Solution:

Step 1:
Use a U-substitution with ${\displaystyle u=3x+2.}$ This means ${\displaystyle du=3dx}$, and after substitution we have
${\displaystyle \int {\frac {1}{3x+2}}=\int {\frac {1}{3u}}du}$
Step 2:
We can now take the integral remembering the special rule:
${\displaystyle \int {\frac {1}{3u}}du={\frac {\log(u)}{3}}}$
Step 3:
Since this integral is an indefinite integral we have to remember to add C at the end.
${\displaystyle \int {\frac {1}{3x+2}}dx={\frac {\ln(3x+2)}{3}}}$