$\int {{{\tan }^{ - 1}}\sqrt {\frac{{1 - \cos 2x}}{{1 + \cos 2x}}} } \;dx = $

$\int \sec^2 x \csc^2 x \, dx = $ . . . . . . $+ C$

$\int \frac{x - 2}{x^2 - 4x + 3} dx = $

Consider the following Assertion $(A)$ and Reason $(R)$:
Assertion $(A)$: $\int \sqrt{x-3} \left(\sin^{-1}(\log x) + \cos^{-1}(\log x)\right) dx = \frac{\pi}{3}(x-3)^{3/2} + c$
Reason $(R)$: $\sin^{-1}(f(x)) + \cos^{-1}(f(x)) = \frac{\pi}{2}$ for $|f(x)| \le 1$
Choose the correct option:

Find the following integral: $\int x^{2}\left(1-\frac{1}{x^{2}}\right) d x$

$\int x(\tan^2 x) dx =$