# lnx Integral

• Dec 9th 2007, 08:20 AM
akhayoon
lnx Integral
$\int \frac{ln|x|} {x\sqrt{1+ln|x|}}$?
• Dec 9th 2007, 08:23 AM
ThePerfectHacker
Let $t=\ln x$ ---> trivial excercise.
• Dec 9th 2007, 08:27 AM
akhayoon
I don't get it
• Dec 9th 2007, 10:40 AM
Krizalid
If $t=\ln x\implies dt=\frac1x\,dx.$

It's just substitution.

Can you handle it from there?
• Dec 9th 2007, 11:26 AM
akhayoon
but then I get

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

how do I integrate this?
• Dec 9th 2007, 11:31 AM
Krizalid
Substitute $\varphi ^2 = 1 + u,$

$\int {\frac{u}
{{\sqrt {1 + u} }}\,du} = 2\int {\left( {\varphi ^2 - 1} \right)\,d\varphi } .$
• Dec 9th 2007, 11:42 AM
akhayoon
Quote:

Originally Posted by Krizalid
Substitute $\varphi ^2 = 1 + u,$

$\int {\frac{u}
{{\sqrt {1 + u} }}\,du} = 2\int {\left( {\varphi ^2 - 1} \right)\,d\varphi } .$

$\int\frac{\varphi^{2}-1}{\varphi}~d\varphi$

so to get ur answer we had to turn $\frac{1}{\varphi}$ into 2?

how does that work

I'm sorry but I'm just trying to understand the solution(Angel)
• Dec 9th 2007, 12:14 PM
Jhevon
Quote:

Originally Posted by akhayoon
$\int\frac{\varphi^{2}-1}{\varphi}~d\varphi$

so to get ur answer we had to turn $\frac{1}{\varphi}$ into 2?

how does that work

I'm sorry but I'm just trying to understand the solution(Angel)

Krizalid thinks a few levels above the average person when it comes to integrals. here is a more conventional substitution for mortals.

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

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

$\Rightarrow dt = du$

So our integral becomes:

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

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