# More integral help!

• January 17th 2011, 02:39 PM
shaunthepanda
More integral help!
So I've been trying to figure out the best way to approach this problem, but so far no luck.

$(int(x*ln(x))/sqrt(x^2-1)$

I'm thinking this has something to do with inverse sin, but I am not sure what to substitute...

• January 17th 2011, 02:46 PM
Jester
if $u = \sqrt{x^2-1}$ then $du = \dfrac{x\, dx}{\sqrt{x^2-1}}$ giving

$\displaystyle \dfrac{1}{2}\int \ln \left(u^2 + 1\right)\, du$ then try parts.
• January 17th 2011, 02:46 PM
dwsmith
Quote:

Originally Posted by shaunthepanda
So I've been trying to figure out the best way to approach this problem, but so far no luck.

$(int(x*ln(x))/sqrt(x^2-1)$

I'm thinking this has something to do with inverse sin, but I am not sure what to substitute...

Integration by parts

$u = ln(x)$

$dv = \frac{1}{\sqrt{x^2-1}}$
• January 17th 2011, 03:24 PM
shaunthepanda
Ok so if u=ln(x) then dv should equal x/sqrt(x^2-1) but if that is the case then how do I integrate x/sqrt(x^2-1)?
• January 17th 2011, 03:55 PM
skeeter
Quote:

Originally Posted by shaunthepanda
Ok so if u=ln(x) then dv should equal x/sqrt(x^2-1) but if that is the case then how do I integrate x/sqrt(x^2-1)?

$\displaystyle \int \frac{x}{\sqrt{x^2-1}} \, dx = \sqrt{x^2-1} + C$
• January 17th 2011, 04:58 PM
dwsmith
Quote:

Originally Posted by shaunthepanda
Ok so if u=ln(x) then dv should equal x/sqrt(x^2-1) but if that is the case then how do I integrate x/sqrt(x^2-1)?

Here is what I was thinking.

$\displaystyle u=\ln{x} \ \ \ \ dv=\frac{1}{\sqrt{x^2-1}}$

$\displaystyle du=\frac{dx}{x} \ \ \ \ v=\ln{(\sqrt{x^2-1}+x)}$

$\displaystyle uv-\int vdu=\ln{(x)}\cdot\ln{(\sqrt{x^2-1}+x)}-\int\frac{\ln{(\sqrt{x^2-1}+x)}}{x}dx$
• January 18th 2011, 07:17 AM
HallsofIvy
Quote:

Originally Posted by dwsmith
Here is what I was thinking.

$\displaystyle u=\ln{x} \ \ \ \ dv=\frac{1}{\sqrt{x^2-1}}$

$\displaystyle du=\frac{dx}{x} \ \ \ \ v=\ln{(\sqrt{x^2-1}+x)}$

$\displaystyle uv-\int vdu=\ln{(x)}\cdot\ln{(\sqrt{x^2-1}+x)}-\int\frac{\ln{(\sqrt{x^2-1}+x)}}{x}dx$

You seem to have neglected the first "x" in the integral:
$\int x\frac{ln(x}}{\sqrt{x^2- 1}}dx$
• January 18th 2011, 07:21 AM
Prove It
Shouldn't it be $\displaystyle u = \ln{x}$ and $\displaystyle dv = \frac{x}{\sqrt{x^2 - 1}}\,dx$?
• January 18th 2011, 10:56 AM
dwsmith
Quote:

Originally Posted by Prove It
Shouldn't it be $\displaystyle u = \ln{x}$ and $\displaystyle dv = \frac{x}{\sqrt{x^2 - 1}}\,dx$?

Quote:

Originally Posted by HallsofIvy
You seem to have neglected the first "x" in the integral:
$\int x\frac{ln(x}}{\sqrt{x^2- 1}}dx$

Try my integration and with the x. They should yield the same results.