A) \[{{\cos }^{-1}}(\log x)+c\]
B) \[x\log (1-{{x}^{2}})+c\]
C) \[{{\sin }^{-1}}(\log x)+c\]
D) \[\frac{1}{2}{{\cos }^{-1}}(\log x)+c\]
Correct Answer: C
Solution :
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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