Q65 / 200 · Manipal (MET) 2011 · Mathematics

The value of $\int \frac{d x}{x \sqrt{1-(\log x)^2}}$ is

  • A
    $\cos ^{-1}(\log x+C)$
  • B
    $x \log \left(1-x^2\right)+C$
  • C
    $\frac{1}{2} \cos ^{-1}(\log x+C)$
  • Correct answer
    D
    $\sin ^{-1}(\log x)+C$

Solution

$\begin{aligned} & \text { Let } I=\int \frac{d x}{x \sqrt{1-(\log x)^2}} \\ & \text { Put } \log x=t \\ & \Rightarrow \quad \frac{1}{x} d x=d t \\ & \begin{aligned} & \therefore I=\int \frac{d t}{\sqrt{1-t^2}}=\sin ^{-1} t+C \\ &=\sin ^{-1}(\log x)+C\end{aligned}\end{aligned}$

Hence, the correct answer is option: 'D'.

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