# Thread: Help with an indefinite integral.

1. ## Help with an indefinite integral.

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

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

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

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

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

So now I have

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

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

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

Did I do my math correctly?

2. Originally Posted by xwanderingpoetx
I got an answer for this problem, but I'm just wondering I did it right.

$\int \frac{8ln(3)log_3 x}{x}$
Where is the "dx" ?
I made my $u = log_3x$ and took the derivative.

$dx = \frac{1}{xln(3)du}$
I moved the $ln(3)$ to the other side because I only have a $1/x$ in the original equation.
I did not follow you here, You should multiply the integrated function by $\frac{ln(3)}{ln(3)}$ ; to get the $ln(3)$ in the denominator.
So now I have

$\int 8ln(3)^2udu$
Here we should explain. Actually, the notation $ln(3)^2=(ln(3))^2$. You should have a squared "ln" in the numerator because you multiply by $\frac{ln(3)}{ln(3)}$ and you have $ln(3)$ in the numerator. so, you should have a squared "ln(3)", What did you write here is the right notation for $( ln(3) )^2$, 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:

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

Did I do my math correctly?
hmmmm, seems correct answer for me.

But you can solve it easily. Note that $log_{3}(x)=\frac {ln(x)}{ln(3)}$
If you substitute this in the original integral and the two "ln(3)" will cancel.
and take the "8" before the integral, You will face the following integral:
$8\int \frac{ln(x)}{x} dx$.
Which can be solved by the simple substitution $u=ln(x)$.