# Thread: Integration problem with ln (using parts)

1. ## Integration problem with ln (using parts)

I've never worked with this kind of problem before. Can anyone please provide a step-by-step answer?

2. let $\displaystyle u = (\ln 5x)^{2}$ and $\displaystyle dv = dx$

then $\displaystyle du = 2 \ln (5x)*\frac{5}{x} \ dx$ and $\displaystyle v =x$

$\displaystyle \int (ln 5x)^{2} = x (\ln 5x)^{2} - 10 \int \ln 5x \ dx$

Integrating by parts again, you can show that $\displaystyle \int \ln 5x = \frac{1}{5}x \ln 5x -\frac{1}{5}x +C$

so $\displaystyle \int (ln 5x)^{2} = x (\ln 5x)^{2} - 2x \ln 5x + 2x + C$

3. How were you able to get du?

4. Originally Posted by Archduke01
How were you able to get du?
I used the chain rule.

$\displaystyle \frac{du}{dx} = 2*(\ln 5x)^{1}*\frac{d}{dx} \ \ln 5x = 2(\ln 5x) * \frac{5}{x}$

so $\displaystyle du = 2(\ln 5x) * \frac{5}{x} \ dx$

5. If $\displaystyle \frac {d}{dx} ln 5x = 5/x$ then how did you get $\displaystyle (1/5)x$ in the 2nd integration by parts?