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 $f$ and $g$ are differentiable functions, then
$(f\circ g)'(x)=f'(g(x))\cdot g'(x).$
The Product Rule: If $f$ and $g$ are differentiable functions, then
$(fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).$
The Quotient Rule: If $f$ and $g$ are differentiable functions and $g(x)\neq 0$ , then
$\left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.$
Additionally, we will need our power rule for differentiation:
$\left(x^{n}\right)'\,=\,nx^{n-1},$ for $n\neq 0$,
as well as the derivative of natural log:
$\left(\ln x\right)'\,=\,{\frac {1}{x}}.$

 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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