Integral with inverse trig functions and u substitution I've been trying to find the integral of $\dfrac{dx}{x \sqrt{x^2-4}}$. Currently, following the example problem, I have it as $\dfrac{1}{2} S \dfrac{1\,du}{x \sqrt{4 [(\frac{x}{2})^2-1]}}$. From what I've got it looks like it's going to the formula matching $\sec^{-1}(x)$. But at this point I'm not sure what to do now about the 4 in there that will get me to the $1/2$ coefficient. I have $u = x/2$, and $du=\frac{1}{2} \, dx$. The answer given is $-\frac{1}{2}\csc^{-1}(\frac{x}{2})+C$. The opposite trig function from what I thought and with a negative somehow. 
 A: \begin{align}
& \frac{dx}{x \sqrt{x^2-4}} = \frac{dx}{x\sqrt{4\left(\left(\frac x 2\right)^2 - 1\right)}} = \frac 1 {\sqrt 4} \cdot \frac{dx}{x\sqrt{\left( \frac x 2 \right)^2 - 1}} = \frac 1 2 \cdot \frac{dx/2}{\frac x 2 \sqrt{\left( \frac x 2 \right)^2 - 1}} \\[10pt]
 = {} & \frac 1 2 \cdot \frac{du}{u\sqrt{u^2 -1}} = \frac 1 2 \cdot \frac{\sec\theta\tan\theta \, d\theta}{\sec\theta\sqrt{\sec^2\theta - 1}} = \frac{d\theta} 2.
\end{align}
A: 
$$
\int\frac{dx}{x\sqrt{x^2-4}}
~=~
\int\frac{x\cdot dx}{x^2\sqrt{x^2-4}}
~=~
\frac12\int\frac{d\big(x^2-4\big)}{\Big[\big(x^2-4\big)+4\Big]\sqrt{x^2-4}}
~=~
\frac12\int\frac{dt}{(t+4)\sqrt t}
$$

Now let $t=u^2.~$ :-$)$
A: It looks like the problem is the radical, namely: $\sqrt{x^2 - 4}$. If you make the trig substitution $x = 2\sec \theta$, then $\sqrt{x^2 - 4} = \sqrt{4(\sec^2 \theta - 1)} = \sqrt{4 \tan^2 \theta} = 2\tan \theta$. Similarly, $\frac{dx}{d \theta} = 2\sec \theta \tan \theta$, so $dx = 2\sec \theta \tan \theta \ d\theta$. So, the integral is:
$$\int \frac{2 \sec \theta \tan \theta}{(2\sec \theta)(2\tan \theta)} \ d\theta$$
$$ = \frac{1}{2}\int d\theta$$
