Question:
If \( \int \frac{dx}{\sqrt{e^{-2x} - 1}} \) is equal to :
If \( \int \frac{dx}{\sqrt{e^{-2x} - 1}} \) is equal to :
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Whenever an integral contains terms like \( e^{-x} \) or \( e^{-2x} \) inside a radical or denominator, multiplying the numerator and denominator by a suitable power of \( e^x \) often clears the negative exponents and reveals a straightforward substitution.
Updated On: Jul 4, 2026
- \( \sin^{-1} e^{-x} + C \)
- \( \log|e^{-x} + \sqrt{e^{-2x}-1}| + C \)
- \( \sin^{-1} e^x + C \)
- \( \log|e^{-x} - \sqrt{e^{-2x}-1}| + C \)
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The Correct Option is C
Solution and Explanation
Concept:
To solve the indefinite integral \( I = \int \frac{dx}{\sqrt{e^{-2x} - 1}} \), we need to simplify the integrand using algebraic manipulation of exponents and then apply a suitable substitution that transforms it into a standard integral form, specifically matching the standard form \( \int \frac{du}{\sqrt{1-u^2}} = \sin^{-1}u + C \).
Step 1: Simplify the expression inside the square root.
We can write \( e^{-2x} \) as \( \frac{1}{e^{2x}} \). Let us rewrite the integral: \[ I = \int \frac{dx}{\sqrt{\frac{1}{e^{2x}} - 1}} = \int \frac{dx}{\sqrt{\frac{1 - e^{2x}}{e^{2x}}}} \] Taking \( e^{2x} \) out of the square root in the denominator gives \( e^x \): \[ I = \int \frac{dx}{\frac{\sqrt{1 - e^{2x}}}{e^x}} = \int \frac{e^x dx}{\sqrt{1 - (e^x)^2}} \]
Step 2: Using the substitution method.
Let us substitute \( u = e^x \). Differentiating both sides with respect to \( x \) yields: \[ du = e^x dx \] Substituting these components into our integral expression gives: \[ I = \int \frac{du}{\sqrt{1 - u^2}} \]
Step 3: Integrating using standard formulas.
The integral is now in a standard form whose solution is well known: \[ \int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}(u) + C \] Substituting back the original value of \( u = e^x \): \[ I = \sin^{-1}(e^x) + C \] This perfectly aligns with option (C).
Step 1: Simplify the expression inside the square root.
We can write \( e^{-2x} \) as \( \frac{1}{e^{2x}} \). Let us rewrite the integral: \[ I = \int \frac{dx}{\sqrt{\frac{1}{e^{2x}} - 1}} = \int \frac{dx}{\sqrt{\frac{1 - e^{2x}}{e^{2x}}}} \] Taking \( e^{2x} \) out of the square root in the denominator gives \( e^x \): \[ I = \int \frac{dx}{\frac{\sqrt{1 - e^{2x}}}{e^x}} = \int \frac{e^x dx}{\sqrt{1 - (e^x)^2}} \]
Step 2: Using the substitution method.
Let us substitute \( u = e^x \). Differentiating both sides with respect to \( x \) yields: \[ du = e^x dx \] Substituting these components into our integral expression gives: \[ I = \int \frac{du}{\sqrt{1 - u^2}} \]
Step 3: Integrating using standard formulas.
The integral is now in a standard form whose solution is well known: \[ \int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}(u) + C \] Substituting back the original value of \( u = e^x \): \[ I = \sin^{-1}(e^x) + C \] This perfectly aligns with option (C).
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