In this integral how do they get $$dx=2\sec^2\theta \ d\theta$$ Here is the integral: $$\int\frac{dx}{x^2\sqrt{x^2+4}}=\int\frac{2\sec^2d\theta}{\mathrm{4\tan^2\ \theta}\!\cdot\!\mathrm{2\sec\ \theta}}=\frac{1}{4}\int\frac{sec\ \theta}{tan^2\ \theta} \ d\theta$$
If $$u=tan\ \theta$$ then it follows that $$d\theta=sec^2\ d\theta$$
With the substitution therein lays my confusion.
The correct substitution: $$u=2\tan\ \theta$$
Here,
$ x = 2 \tan θ $ is the substitution, which implies that, $ dx = 2\sec ^2θ .dθ$
The first target here is to get rid of the square root in the denominator, which is the primary reason why $ 2 \tan θ $ is substituted instead of $ \tan θ $.
Now, $$ \int \frac {dx}{x^2 \sqrt {x^2 + 4}}$$ $$ = \int \frac {2\sec^2 θ \,dθ } {4 \tan^2 θ \sqrt{4(\tan^2 θ + 1)}}$$ $$ = \int \frac {2\sec^2θ \, dθ}{4 \tan^2 θ \,2\secθ} $$ $$ = \frac {1}{4} \int \frac {\sec θ}{\tan^2 θ} \,dθ $$ (as you said. Here I am assuming that till now, you have done the integral correctly. If yes, then this is how it is continued:) $$ = \frac {1}{4} \int \operatorname{cosec} \,θ \cot θ \,dθ $$ $$ = \frac {1}{4} (-\sin θ) + c $$ $$ = \frac {-1}{4} \sin (\tan ^{-1}{(x/2))} + c$$ $$ = \frac {-x}{4(\sqrt{x^2 + 4})} + c$$ (In last step, I changed tan inverse to sin inverse to simplify it.)
