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$\int \frac{1}{3x+12} \, dx =$

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Evaluate definite integrals using the power rule

Full solution

Q.

\int \frac{1}{3x+12} \, dx =

Factor out constant: Simplify the integral by factoring out the constant from the denominator. $\newline$ We have: $\int \frac{1}{3x+12}\,dx$ $\newline$ Factor out the $3$ from the denominator: $\int \frac{1}{3(x+4)}\,dx$ $\newline$ This can be rewritten as: $(\frac{1}{3})\int \frac{1}{x+4}\,dx$
Identify natural logarithm function: Identify the integral as a natural logarithm function. $\newline$ The integral of $\frac{1}{x+a}$ with respect to $x$ is $\ln|x+a| + C$ , where $C$ is the constant of integration. $\newline$ So, $(\frac{\(1\)}{\(3\)})\int \frac{\(1\)}{x+\(4\)}\,dx = (\frac{\(1\)}{\(3\)})\ln|x+\(4\)| + C
Write final answer: Write the final answer.\(\newline\)The integral of \(\frac{1}{3x+12}\) with respect to \(x\) is \((\frac{1}{3})\ln|x+4| + C\).

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