Question

The antiderivative of $\dfrac{1}{x\sqrt{x^2-1}},\; x>1$ with respect to $x$ is

• A. $\sin^{-1}x + C$
• B. $\cos^{-1}x + C$
• C. $\sec^{-1}x + C$
• D. $\cot^{-1}x + C$

Hint:

Remember the important identity: $\dfrac{d}{dx}(\sec^{-1}x) = \dfrac{1}{x\sqrt{x^2-1}}$ for $x>1$.

Answer 1

The Correct Option is C

Step 1: Recall the derivative of inverse secant.
From standard differentiation formulas of inverse trigonometric functions, \[ \frac{d}{dx}(\sec^{-1}x) = \frac{1}{x\sqrt{x^2-1}}, \quad x>1 \] This derivative exactly matches the integrand given in the question. 

Step 2: Compare with the given integral. 
The integral provided is \[ \int \frac{1}{x\sqrt{x^2-1}}\,dx \] Since the derivative of $\sec^{-1}x$ equals the given expression, the integral directly becomes \[ \sec^{-1}x + C \] where \(C\) is the constant of integration. 

Step 3: Conclusion. 
Thus the antiderivative of the function is the inverse secant function. 
Final Answer: $\boxed{\sec^{-1}x + C}$