# Integration Problem

• Aug 5th 2010, 05:54 PM
Bushy
Integration Problem
Given

$\displaystyle \int \frac{\ln bx}{x^2} ~dx =-\frac{\ln bx}{x}-\frac{1}{x}+C$

Then find $\displaystyle b$ where

$\displaystyle \int_{\frac{1}{b}}^4 \frac{5\ln bx}{x^2} ~dx = 5$

I'm saying that

$\displaystyle 5\int_{\frac{1}{b}}^4 \frac{\ln bx}{x^2} ~dx = 5$

Dividing both sides by $\displaystyle 5$

$\displaystyle \int_{\frac{1}{b}}^4 \frac{\ln bx}{x^2} ~dx = 1$

$\displaystyle \left[ -\frac{\ln bx}{x}-\frac{1}{x}\right]_{\frac{1}{b}}^4 = 1$

$\displaystyle \left[ -\frac{\ln 4b}{4}-\frac{1}{4}\right]-\left[ -\frac{\ln b\frac{1}{b}}{\frac{1}{b}}-\frac{1}{\frac{1}{b}}\right] = 1$

$\displaystyle \left[ -\frac{\ln 4b}{4}-\frac{1}{4}\right]-\left[ -0-b\right] = 1$

$\displaystyle -\frac{\ln 4b}{4}-\frac{1}{4}+b = 1$

Here is where I am stuck because I can't find an explicit way to solve for $\displaystyle b$ Am I missing a log law or something similar?

Thanks..
• Aug 5th 2010, 05:56 PM
Also sprach Zarathustra
Everything is correct!

You can't solve it for b.
• Aug 5th 2010, 06:12 PM
Bushy
Quote:

Originally Posted by Also sprach Zarathustra
Mistake at the third line.

Quote:

$\displaystyle \int_{\frac{1}{b}}^4 \frac{5\ln bx}{x^2} ~dx = 5$

I'm saying that

$\displaystyle 5\int_{\frac{1}{b}}^4 \frac{\ln bx}{x^2} ~dx = 5$

Dividing both sides by $\displaystyle 5$

I can't take a constant out?
• Aug 5th 2010, 06:23 PM
Also sprach Zarathustra
All this is after you fixing your mistake...
• Aug 5th 2010, 07:09 PM
Bushy
I did an edit, sorry about that.

Do you think I have solved the problem ok, in light of the edit?
• Aug 5th 2010, 08:56 PM
Bushy
I suppose the question is, is the method correct?
• Aug 5th 2010, 09:00 PM
Also sprach Zarathustra
YES...