Given: $\large \int \frac {1 + \tan^2 x} {\sqrt (1 - \tan^2 x)}\;$$dx$
Let $\tan x = t \rightarrow \sec^2x\;dx = dt$
We know that: $1 + \tan^2 x = \sec^2 x$
Substituting, we get: $\large \int \frac {1 + \tan^2 x} {\sqrt (1 - \tan^2 x)}\;$$dx = \large \int \frac {\sec^2x\;dx}{\sqrt(1 - \tan^2x)}$
$\Rightarrow \large \int \frac {1 + \tan^2 x} {\sqrt (1 - \tan^2 x)}\;$$dx = \large \int \frac{dt}{\sqrt (1-t^2)}$
We know that: $\large \int \frac{1}{\sqrt(1-x^2)}\;$$dx =\sin^{-1} x + k$
$\Rightarrow \large \int \frac {1 + \tan^2 x} {\sqrt (1 - \tan^2 x)}\;$$dx = \sin^{-1}(t) + k$
$\Rightarrow \large \int \frac {1 + \tan^2 x} {\sqrt (1 - \tan^2 x)}\;$$dx = \sin^{-1}(\tan x) + k$