Indefinite Integral?

[sausu]

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

It's some secant inverse...Here's what I got so far:

Int(1/(x(Sqrt(4(x^2 /4)-1)
Int(1/(2x(Sqrt((x^2 /4)-1)
1/2 int(1/(x(Sqrt((x /2)^2)-1)
U-sub
u=x/2
du=1/2 dx
2du= dx
(This is where I hit a wall..I have no clue what I'm doing)

Please help me and show step by step.

 

[galactus]

There are various ways to tackle it. Do you have to use a certain method?.

Trig sub works well with this one.

\(\displaystyle \int\frac{1}{x\sqrt{x^{2}-4}}dx\)

Let \(\displaystyle x=2sec(t), \;\ dx=2sec(t)tan(t)dt\)

Make the subs:

\(\displaystyle \int\frac{2sec(t)tan(t)}{2sec(t)\sqrt{(2sec(t))^{2}-4}}dt\)

\(\displaystyle \int\frac{tan(t)}{\sqrt{4(sec^{2}(t)-1)}}dt\)

Remember the identity \(\displaystyle tan^{2}(t)=sec^{2}(t)-1\)

\(\displaystyle \frac{1}{2}\int dt=\frac{t}{2}\)

Now, resub. Noting that \(\displaystyle t=sec^{-1}(\frac{x}{2})\)

\(\displaystyle \boxed{\frac{1}{2}sec^{-1}(\frac{x}{2})=\frac{1}{2}cos^{-1}(\frac{2}{x})}\)

You may see this written in some form that looks completely different, but is actually equivalent.

Such as, \(\displaystyle \frac{1}{2}tan^{-1}\left(\frac{\sqrt{x^{2}-4}}{2}\right)\)