# integrating a function

• Feb 10th 2009, 11:34 PM
twilightstr
integrating a function
integrate xlnx/(square root x^2-1)
how i started:
by using integration by parts i got (lnx)(2/3(x^2-1)^(3/2) - integral 2/3(x^2-1)^(3/2)/(x)
where i let u= lnx v=2/3(x^2-1)^3/2
du= (1/x)dx dv= x(x^2-1)^-(1/2)dx
I dont how to take the integral of that function.
• Feb 11th 2009, 01:24 AM
Check the formula you used and than the integration you performed(Thinking) Is it correct(Shake)
• Feb 14th 2009, 07:13 AM
integration by parts means
$\displaystyle \int{u \times v dx} = u\int{vdx}-\int{(u'\int{vdx}) dx}$

So for your function lets take $\displaystyle u= ln(x)$ and $\displaystyle v=\frac{x}{\sqrt{x^2-1}}$

Hence the integral becomes

$\displaystyle \int{\frac{x~ln(x)dx}{\sqrt{x^2-1} } } = ln(x) \int{\frac{xdx}{\sqrt{x^2-1}} }- \int{((\frac{d}{dx} ln(x))\int{\frac{xdx}{\sqrt{x^2-1}}})dx}$

First of all
$\displaystyle \int{\frac{x~dx}{\sqrt{x^2-1}}}$

Put $\displaystyle x^2-1 = t$

Hence $\displaystyle xdx =dt/2$

Hence the integral becomes
$\displaystyle \int{\frac{2~dt}{\sqrt{t}}}$

$\displaystyle =\frac{2\sqrt{t}}{2} + c$

And

$\displaystyle \frac{d}{dx} ln(x) =\frac{1}{x}$

(Nod)