Intergral

• Jan 8th 2008, 05:28 PM
polymerase
Intergral
$\displaystyle \int\frac{x\;lnx}{\sqrt{1-x^2}}\;dx$

Thx
• Jan 8th 2008, 06:02 PM
Jhevon
Quote:

Originally Posted by polymerase
$\displaystyle \int\frac{x\;lnx}{\sqrt{1-x^2}}\;dx$

Thx

can't really see an elegant way to do this one. but integrating by parts should work (you will need to do several substitutions afterwards, ok, like 2, "several" makes it sound like a lot), so yeah, this will be messy.

let $\displaystyle \frac x{\sqrt{1 - x^2}}$ be the functoin you integrate, i think it is convention to call this your $\displaystyle dv$ and let $\displaystyle u$ (the function you differentiate) be $\displaystyle \ln x$

i'm sure one of the best integrator nominees (:D) can find a better way to do this
• Jan 8th 2008, 06:32 PM
mr fantastic
Quote:

Originally Posted by Jhevon
can't really see an elegant way to do this one. but integrating by parts should work (you will need to do several substitutions afterwards, ok, like 2, "several" makes it sound like a lot), so yeah, this will be messy.

let $\displaystyle \frac x{\sqrt{1 - x^2}}$ be the functoin you integrate, i think it is convention to call this your $\displaystyle dv$ and let $\displaystyle u$ (the function you differentiate) be $\displaystyle \ln x$

i'm sure one of the best integrator nominees (:D) can find a better way to do this

Yes, that would work (the technique, not the last line - although that might work too [depending on what you mean by 'better']) and it's pretty routine.

And to solve the resulting integral the substitution $\displaystyle w^2 = 1 - x^2$ would be made, followed by a spot of partial fraction decomposition.
• Jan 8th 2008, 06:38 PM
Jhevon
Quote:

Originally Posted by mr fantastic
Yes, that would work (the technique, not the last line - although that might work too [depending on what you mean by 'better']) and it's pretty routine.

And to solve the resulting integral the substitution $\displaystyle w^2 = 1 - x^2$ would be made, followed by a spot of partial fraction decomposition.

ah, yes! i did it the hard way, which is why i needed more than one substitutions. i let $\displaystyle u = 1 - x^2$, $\displaystyle u^2 = 1 - x^2$ does make more sense