$\displaystyle \int \frac{ln|x|} {x\sqrt{1+ln|x|}} $?

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- Dec 9th 2007, 07:20 AMakhayoonlnx Integral
$\displaystyle \int \frac{ln|x|} {x\sqrt{1+ln|x|}} $?

- Dec 9th 2007, 07:23 AMThePerfectHacker
Let $\displaystyle t=\ln x$ ---> trivial excercise.

- Dec 9th 2007, 07:27 AMakhayoon
I don't get it

- Dec 9th 2007, 09:40 AMKrizalid
If $\displaystyle t=\ln x\implies dt=\frac1x\,dx.$

It's just substitution.

Can you handle it from there? - Dec 9th 2007, 10:26 AMakhayoon
but then I get

$\displaystyle \int\frac{u}{\sqrt{1+u}}~dt$

how do I integrate this? - Dec 9th 2007, 10:31 AMKrizalid
Substitute $\displaystyle \varphi ^2 = 1 + u,$

$\displaystyle \int {\frac{u}

{{\sqrt {1 + u} }}\,du} = 2\int {\left( {\varphi ^2 - 1} \right)\,d\varphi } .$ - Dec 9th 2007, 10:42 AMakhayoon
- Dec 9th 2007, 11:14 AMJhevon
Krizalid thinks a few levels above the average person when it comes to integrals. here is a more conventional substitution for mortals.

$\displaystyle \int \frac u{\sqrt{1 + u}}~du$

Let $\displaystyle t = 1 + u \implies \boxed{u = t - 1}$

$\displaystyle \Rightarrow dt = du$

So our integral becomes:

$\displaystyle \int \frac {t - 1}{\sqrt{t}}~dt$

which we can do using the power rule (after dividing $\displaystyle \sqrt{t}$ into each of the terms in the numerator)