# 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
$
\int{u \times v dx} = u\int{vdx}-\int{(u'\int{vdx}) dx}

$

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

$

Hence the integral becomes

$

\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
$
\int{\frac{x~dx}{\sqrt{x^2-1}}}

$

Put $x^2-1 = t$

Hence $xdx =dt/2$

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

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

And

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

(Nod)