Question

\(\int \sqrt{x^2-6 x-16} \mathrm{~d} x\) equals

• A. \((\frac{x-3}{2})\sqrt{x^{2}-6x-16}+\frac{5}{2}\mathrm{log}(x-3+\sqrt{x^{2}-6x-16})+c\) where c is the constant of integration
• B. \(\begin{array}{cc} & (\frac{x-3}{2})\sqrt{x^{2}-6x-16}-\frac{25}{2}\mathrm{log}(x-3+\sqrt{x^{2}-6x-16})+c \end{array}\) where c is the constant of integration
• C. \((\frac{x-3}{2})\sqrt{x^{2}-6x-16}+\frac{25}{2}\mathrm{log}(x-3+\sqrt{x^{2}-6x-16})+c\) where c is the constant of integration
• D. \(\begin{array}{cc} (\frac{x-3}{2})\sqrt{x^{2}-6x-16} & -\frac{5}{2}\mathrm{log}(x-3+\sqrt{x^{2}-6x-16})+c \end{array}\) where \(c\) is the constant of integration

Hint:

Complete the square and use the standard formula for the integral of the square root of x squared minus a squared.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Complete the square: \(x^{2}-6x-16=(x-3)^{2}-25\), so the integral is of the form \(\int\sqrt{u^{2}-a^{2}}\,du\) with \(u=x-3,\ a=5\).

Step 2: Use the Standard Formula
\(\int\sqrt{u^{2}-a^{2}}\,du=\dfrac{u}{2}\sqrt{u^{2}-a^{2}}-\dfrac{a^{2}}{2}\log\left|u+\sqrt{u^{2}-a^{2}}\right|+c\).

Step 3: Substitute
With \(u=x-3\) and \(a^{2}=25\): \(\dfrac{x-3}{2}\sqrt{x^{2}-6x-16}-\dfrac{25}{2}\log\left(x-3+\sqrt{x^{2}-6x-16}\right)+c\).

Step 4: Match
The minus sign and the coefficient \(\dfrac{25}{2}\) match option (B).

Final Answer:
The integral equals the expression with \(-\dfrac{25}{2}\log\) in option (B). \[ \boxed{\text{(B) }\left(\dfrac{x-3}{2}\right)\sqrt{x^{2}-6x-16}-\dfrac{25}{2}\log\left(x-3+\sqrt{x^{2}-6x-16}\right)+c} \]