How to integrate $\int {2\over (x^2+2)\sqrt{x^2+4}}dx$? 
Solve the following indefinite integral:
$$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx$$

My approach:
I used the substitution: $x=2\tan t$, $dx=2\sec^2t dt$
$$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx=\int \frac{2}{(4\tan^2t+2)\sqrt{4\tan^2t+4}}\cdot 2\sec^2t\ dt$$
$$=\int \frac{4\sec^2t }{2(2\tan^2t+1)2\sec t} dt$$
$$=\int \frac{\sec t}{2\tan^2t+1}dt$$
In numerator I have $\sec t$ but not $\sec^2t$ therefore I can't see a way to take it further. Please help me solve this integral. Thanks in advance.

Alternatively, let $ x = 2\sinh t \Rightarrow \frac{dx}{dt} = 2\cosh t$.
The integral becomes
$$\begin{array} {r c l }
\displaystyle \int \frac2{(4\sinh^2 t + 2)(2\cosh t)} \cdot 2\cosh t \, dt 
&=& \displaystyle \frac12 \int \frac1{2\sinh^2 t + 1} \, dt \\
&=& \displaystyle \frac12 \tan^{-1} (\tanh t) + C \\
&=& \displaystyle \frac12 \tan^{-1} \left (\tanh \left (\sinh^{-1} \tfrac x2 \right )\right) + C \\
\end{array} $$
