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I believe this should be a simple problem but I don't have an answer key to confirm if this is right, and some of the similar questions I can find online seem to be giving more complicated solutions.

The problem is: $ \int \frac{ln 3x}{x} dx $

And this is my solution, based on the fact that $ \int ln(ax) dx = \frac{1}{x} $:

$$ u = ln(3x) $$ $$ \implies \frac{du}{dx} = \frac{1}{x} $$ $$ \implies du = \frac{1}{x} dx $$ So: $$ \int \frac{ln 3x}{x} dx = \int u . du$$ $$ \implies \frac{u^2}{2} + C$$ $$ \implies \frac{(ln 3x)^2}{2} + C$$

Is that right?

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    $\begingroup$ Yes, it is.${}$ $\endgroup$ – David Mitra Apr 20 '15 at 8:17
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    $\begingroup$ Yes it is correct. A quick check is to take the derivative of your answer and see you get the expression inside the integral $\endgroup$ – Brenton Apr 20 '15 at 8:18
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One way to check the result. Since $$\int\frac{\ln(3x)}{x}dx=\frac{(\ln(3x))^2}{2}+C$$ then $$\frac{d}{dx}\frac{(\ln(3x))^2}{2}+C$$ should be the initial function. Hence, doing the differentiation \begin{align} \frac{1}{2}\frac{d}{dx}(\ln(3x))^2&=\ln(3x)\frac{d}{dx}\ln(3x)\\ &=\ln(3x)\cdot\frac{1}{3x}\cdot 3\\ &=\frac{\ln(3x)}{x}. \end{align} This is your initial function, and hence your integration was correct.

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