Integral of ln (3x) / x

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?

• Yes, it is.${}$ – David Mitra Apr 20 '15 at 8:17
• Yes it is correct. A quick check is to take the derivative of your answer and see you get the expression inside the integral – Brenton Apr 20 '15 at 8:18

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.