$\int \large\frac{1}{x} . \sqrt {\large\frac{1-x}{1+x} \times \frac{1-x}{1-x}}$$dx$
$\qquad= \int \large\frac{1}{x} \frac{(1-x)}{\sqrt {1-x^2}}$$dx$
$\qquad= \int \large\frac{1}{x \sqrt {1-x^2}} \frac{1}{\sqrt {1-x^2}}$$dx$
$\qquad= \int \large\frac{1}{x \sqrt {1-x^2}} $$dx-\int \frac{1}{\sqrt {1-x^2}}$$dx$
$\qquad= \int \large\frac{1}{x \sqrt {1-x^2}}$$dx-\sin ^{-1} x +c$
$x= \sin \theta $
differentiate with x
$dx= \cos \theta d \theta$
=>$\int \large\frac{\cos \theta d\theta }{\sin \theta . \cos \theta }-$$ \sin ^{-1} x +c$
=>$\int cosec \theta d\theta - \sin ^{-1} x +c$
=> $\log | cosec \theta - \cot \theta | - \sin ^{-1} x +c$
$\log |\large\frac{1-\sqrt {1-x^2}}{x}|$$-\sin ^{-1}+c$
Hence a is the correct answer.