Find the antiderivative of 
| Foundations:
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| This problem requires two rules of integration. In particular, you need
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Integration by substitution (u - sub): If  is a differentiable functions whose range is in the domain of 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 f}
, then
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- 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 g'(x)f(g(x)) dx \,=\, \int f(u) du.}
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| We also need the derivative of the natural log since we will recover natural log from integration:
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- 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 \left(ln(x)\right)' \,=\, \frac{1}{x}.}
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Solution:
| Step 1:
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Use a u-substitution with 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.}
This means 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 du = 3\,dx}
, or 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 dx=du/3}
. After substitution we have
- 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 \,=\, \int \frac{1}{u}\,\frac{du}{3}\,=\,\frac{1}{3}\int\frac{1}{u}\,du.}
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| Step 2:
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| We can now take the integral remembering the special rule:
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- 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{1}{3}\int\frac{1}{u}\,du\,=\, \frac{\log(u)}{3}.}
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| Step 3:
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Now we need to substitute back into our original variables using our original substitution 
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| to find 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}.}
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| Step 4:
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| Since this integral is an indefinite integral we have to remember to add a constant 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 C}
at the end.
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| Final Answer:
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- 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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