Originally Posted by

**xwanderingpoetx** I got an answer for this problem, but I'm just wondering I did it right.

$\displaystyle \int \frac{8ln(3)log_3 x}{x}$

**Where is the "dx" ?**

I made my $\displaystyle u = log_3x$ and took the derivative.

$\displaystyle dx = \frac{1}{xln(3)du}$

I moved the $\displaystyle ln(3)$ to the other side because I only have a $\displaystyle 1/x$ in the original equation.

**I did not follow you here, You should multiply the integrated function by** $\displaystyle \frac{ln(3)}{ln(3)}$ **; to get the** $\displaystyle ln(3)$ **in the denominator.**

So now I have

$\displaystyle \int 8ln(3)^2udu$

**Here we should explain. Actually, the notation $\displaystyle ln(3)^2=(ln(3))^2$. You should have a squared "ln" in the numerator because you multiply by $\displaystyle \frac{ln(3)}{ln(3)}$ and you have $\displaystyle ln(3)$ in the numerator. so, you should have a squared "ln(3)", What did you write here is the right notation for $\displaystyle ( ln(3) )^2$, But clearly, you did not mean that; because you wrtie it as "ln(3)^2".**

So I integrate and re-substitute and I wind up with:

$\displaystyle 4ln(3)^2(log_3x)^2+c$

Did I do my math correctly?

**hmmmm, seems correct answer for me.**