A) \[{{\cos }^{-1}}(\log x+C)\]
B) \[x\log (1-{{x}^{2}})+C\]
C) \[\frac{1}{2}{{\cos }^{-1}}(\log x+C)\]
D) \[{{\sin }^{-1}}(\log x+C)\]
Correct Answer: D
Solution :
Let\[I=\int{\frac{dx}{x\sqrt{1-{{(\log x)}^{2}}}}}\] Put \[\log x=t\] \[\Rightarrow \] \[\frac{1}{x}dx=dt\] \[\therefore \] \[I=\int{\frac{dt}{\sqrt{1-{{t}^{2}}}}={{\sin }^{-1}}t+C}\] \[={{\sin }^{-1}}(\log x)+C\]
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