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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = 3x + 2}
to get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 "+ C" at the end.

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{1}{3x + 2} dx = \frac{\ln(3x + 2)}{3} + C}

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