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?
$\endgroup$ 21 Answer
$\begingroup$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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