_ ^ dx x 1 - ( x) ^2 =

MCQ

$\int_{}^{} {\frac{{dx}}{{x\sqrt {1 - {{(\log x)}^2}} }} = } $

A
${\cos ^{ - 1}}(\log x) + c$
B
$x\log (1 - {x^2}) + c$
✓
${\sin ^{ - 1}}(\log x) + c$
D
$\frac{1}{2}{\cos ^{ - 1}}(\log x) + c$

✓

Answer

Correct option: C.

${\sin ^{ - 1}}(\log x) + c$

c
(c) Put $\log x = t \Rightarrow \frac{1}{x}\,dx = dt,$ then it reduces to
$\int_{}^{} {\frac{{dt}}{{\sqrt {1 - {t^2}} }} = {{\sin }^{ - 1}}t = {{\sin }^{ - 1}}(\log x) + c.} $

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