Question:
The value of
\[
\int \frac{dx}{x \sqrt{1 - (\log x)^2}}
\]
is:
The value of
\[
\int \frac{dx}{x \sqrt{1 - (\log x)^2}}
\]
is:
Show Hint
For integrals involving \( \sqrt{1 - u^2} \), recognize the standard result for the inverse sine function.
Updated On: Apr 22, 2026
- \( \cos^{-1} (\log x + C) \)
- \( x \log (1 - x^2) + C \)
- \( \frac{1}{2} \cos^{-1} (\log x) + C \)
- \( \sin^{-1} (\log x) + C \)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Substitution to simplify the integral.
We are given the integral: \[ I = \int \frac{dx}{x \sqrt{1 - (\log x)^2}} \] Let us make the substitution \( u = \log x \). This implies: \[ du = \frac{dx}{x} \] Therefore, the integral becomes: \[ I = \int \frac{du}{\sqrt{1 - u^2}} \]
Step 2: Recognizing the standard integral form.
We recognize that the integral: \[ \int \frac{du}{\sqrt{1 - u^2}} \] is a standard integral whose result is the inverse sine function \( \sin^{-1}(u) \). Therefore, we have: \[ I = \sin^{-1}(u) + C \]
Step 3: Substituting back \( u = \log x \).
Now, substitute \( u = \log x \) back into the equation: \[ I = \sin^{-1} (\log x) + C \]
Step 4: Conclusion.
Thus, the value of the integral is: \[ \boxed{\sin^{-1} (\log x) + C} \] This matches option (D), which is the correct answer.
We are given the integral: \[ I = \int \frac{dx}{x \sqrt{1 - (\log x)^2}} \] Let us make the substitution \( u = \log x \). This implies: \[ du = \frac{dx}{x} \] Therefore, the integral becomes: \[ I = \int \frac{du}{\sqrt{1 - u^2}} \]
Step 2: Recognizing the standard integral form.
We recognize that the integral: \[ \int \frac{du}{\sqrt{1 - u^2}} \] is a standard integral whose result is the inverse sine function \( \sin^{-1}(u) \). Therefore, we have: \[ I = \sin^{-1}(u) + C \]
Step 3: Substituting back \( u = \log x \).
Now, substitute \( u = \log x \) back into the equation: \[ I = \sin^{-1} (\log x) + C \]
Step 4: Conclusion.
Thus, the value of the integral is: \[ \boxed{\sin^{-1} (\log x) + C} \] This matches option (D), which is the correct answer.
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