Question

The value of \(\int \frac{2x - 9}{\sqrt{x^2 - 9x + 1}} \, dx\) is:

• A. \(2\sqrt{x^2 - 9x + 1} + c\)
• B. \(\frac{1}{2}\sqrt{x^2 - 9x + 1} + c\)
• C. \(\sqrt{x^2 - 9x + 1} + c\)
• D. \(\frac{2}{9}\sqrt{x^2 - 9x + 1} + c\)

Hint:

Exam Tip:
When the numerator is the derivative of the expression under the square root, use substitution \(u =\) the expression under the square root.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to evaluate an integral where the numerator is the derivative of the expression inside the square root (up to a constant factor).

Step 2: Key Formula or Approach:
Notice that \(\frac{d}{dx}(x^2 - 9x + 1) = 2x - 9\).
So, the numerator is exactly the derivative of the denominator's inner function.

Step 3: Detailed Explanation:
Let \(u = x^2 - 9x + 1\). Then \(du = (2x - 9) dx\).
Thus, \[ \int \frac{2x - 9}{\sqrt{x^2 - 9x + 1}} \, dx = \int \frac{du}{\sqrt{u}} = \int u^{-1/2} du = 2u^{1/2} + c = 2\sqrt{x^2 - 9x + 1} + c \] So, the integral is \(2\sqrt{x^2 - 9x + 1} + c\).

Step 4: Final Answer:
This matches option (A). Therefore, option (A) is correct.