Question

Evaluate the indefinite integral $\int \sec^{-1}x \, dx$

• A. $x \sec^{-1}x + \log|x + \sqrt{x^2 - 1}| + c$
• B. $x \sec^{-1}x - \log|x + \sqrt{x^2 - 1}| + c$
• C. $x \sec^{-1}x - \log|x + \sqrt{x^2 + 1}| + c$
• D. $x \sec^{-1}x + \log|x + \sqrt{x^2 + 1}| + c$

Hint:

Be extremely careful with sign signs and terms under the square root radical! Since $\sec^2\theta - 1 = \tan^2\theta$, the core function structure must contain a minus sign inside the radical ($x^2 - 1$). This lets you instantly eliminate choices (C) and (D)!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for the standard antiderivative of the inverse secant function $\sec^{-1}x$.

Step 2: Key Formula or Approach:
To integrate a single isolated inverse trigonometric function, we treat it as a product with $1$ and apply the Integration by Parts method:
$$\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$$ Following the LIATE rule, we select:
$u = \sec^{-1}x \implies \frac{du}{dx} = \frac{1}{x\sqrt{x^2 - 1}}$
$v = 1 \implies \int v \, dx = x$

Step 3: Detailed Explanation:
Let's plug these parts into the integration by parts formula:
$$I = \int \sec^{-1}x \cdot 1 \, dx$$ $$I = (\sec^{-1}x \cdot x) - \int \left( \frac{1}{x\sqrt{x^2 - 1}} \cdot x \right) dx$$ The $x$ terms in the numerator and denominator of the integral cancel out perfectly:
$$I = x \sec^{-1}x - \int \frac{1}{\sqrt{x^2 - 1}} \, dx$$ Now, apply the standard standard formula for the remaining integral, which is $\int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \log|x + \sqrt{x^2 - a^2}|$:
$$\int \frac{1}{\sqrt{x^2 - 1}} \, dx = \log|x + \sqrt{x^2 - 1}|$$ Combining our terms and adding the constant of integration $c$ yields:
$$I = x \sec^{-1}x - \log|x + \sqrt{x^2 - 1}| + c$$ This matches option (B).

Step 4: Final Answer:
The value of the integral is $x \sec^{-1}x - \log|x + \sqrt{x^2 - 1}| + c$, which corresponds to option (B).