# Integration problem with ln (using parts)

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• Mar 10th 2010, 07:37 PM
Archduke01
Integration problem with ln (using parts)
http://euler.vaniercollege.qc.ca/web...fe1e600c91.png

I've never worked with this kind of problem before. Can anyone please provide a step-by-step answer?
• Mar 10th 2010, 08:20 PM
Random Variable
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 $\displaystyle \int (ln 5x)^{2} = x (\ln 5x)^{2} - 2x \ln 5x + 2x + C$
• Mar 11th 2010, 10:13 AM
Archduke01
How were you able to get du?
• Mar 11th 2010, 10:36 AM
Random Variable
Quote:

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 $\displaystyle du = 2(\ln 5x) * \frac{5}{x} \ dx$
• Mar 11th 2010, 03:05 PM
Archduke01
If $\displaystyle \frac {d}{dx} ln 5x = 5/x$ then how did you get $\displaystyle (1/5)x$ in the 2nd integration by parts?