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: Treat it as a product.
Write $\displaystyle\int\sec^{-1}x\,dx=\int \sec^{-1}x\cdot1\,dx$ so we can use integration by parts.
Step 2: Choose the parts.
By LIATE take $u=\sec^{-1}x$ and $dv=1\,dx$, so $du=\dfrac{1}{x\sqrt{x^2-1}}\,dx$ and $v=x$.
Step 3: Apply the formula.
$\displaystyle\int u\,dv=uv-\int v\,du=x\sec^{-1}x-\int x\cdot\dfrac{1}{x\sqrt{x^2-1}}\,dx$.
Step 4: Cancel the $x$.
The remaining integral is $\displaystyle\int\dfrac{dx}{\sqrt{x^2-1}}$.
Step 5: Use the standard integral.
$\displaystyle\int\dfrac{dx}{\sqrt{x^2-1}}=\log\left|x+\sqrt{x^2-1}\right|$.
Step 6: Combine the pieces.
Therefore $\displaystyle\int\sec^{-1}x\,dx=x\sec^{-1}x-\log\left|x+\sqrt{x^2-1}\right|+c$, matching option (2).
\[ \boxed{x\sec^{-1}x-\log\left|x+\sqrt{x^2-1}\right|+c} \]