Integral resulting in an Arc trig function

[Guest]

integral of 1/(x(sqrt x^2-4))dx

by using x^2 as u, du/2=xdx

I end up with 1/4Arcsec [abs(x^2)]/2+C

Once again, my TI89 gives me a different answer. Can anyone help with this? I'd like to know if I did it right and if not, what did I do wrong.

Thanks

 

[pka]

This what I get:
\(\displaystyle \L\int {\frac{{dx}}{{x\sqrt {x^2 - 4} }} = \frac{{ - 1}}{2}\arctan \left( {\frac{2}{{\sqrt {x^2 - 4} }}} \right)} .\)

 

[Guest]

okay, when i did all my work again, I got

1/2Arcsec abs(x)/2+C

if i make u=x and du=dx I can follow the Arcsec formula for integration right?

{integral of du/u sqrt(u^2-a^2) dx = 1/a Arcsec abs(u)/a}

how are you getting to the arctan? Doesn't that apply to a problem like du/a^2+u^2?

 

[soroban]

Hello, ezrajoelmicah!

There is a standard formula for this one.

If you don't know it, use Trig Substitution . . .

\(\displaystyle \L\int\frac{dx}{x\sqrt{x^2\,-\,4}}\)

Let \(\displaystyle \,x\:=\:2\sec\theta\;\;\Rightarrow\;\;dx\:=\:2\sec\theta\tan\theta\,d\theta\)

Substitute: \(\displaystyle \L\L\int\frac{2\sec\theta\tan\theta\,d\theta}{2\sec\theta\cdot 2\tan\theta} \;=\;\frac{1}{2}\int d\theta \;=\;\frac{1}{2}\theta\,+\,C\)


Back-substitute: \(\displaystyle \L\:\frac{1}{2}\,\text{arcsec}\left(\frac{x}{2}\right)\,+\,C\)