Integration Problem

**Bushy** · August 5th 2010, 05:54 PM

Given

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

Then find $b$ where

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

I'm saying that

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

Dividing both sides by $5$

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

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

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

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

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

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

Thanks..

**Also sprach Zarathustra** · August 5th 2010, 05:56 PM

Everything is correct!

You can't solve it for b.

**Bushy** · August 5th 2010, 06:12 PM

Originally Posted by Also sprach Zarathustra

Mistake at the third line.

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

I'm saying that

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

Dividing both sides by $5$

I can't take a constant out?

**Also sprach Zarathustra** · August 5th 2010, 06:23 PM

All this is after you fixing your mistake...

**Bushy** · August 5th 2010, 07:09 PM

I did an edit, sorry about that.

Do you think I have solved the problem ok, in light of the edit?

**Bushy** · August 5th 2010, 08:56 PM

I suppose the question is, is the method correct?

**Also sprach Zarathustra** · August 5th 2010, 09:00 PM

YES...

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