Question

\(\int \frac{x-3}{x^2-9} \, dx\)

• A. \( \log(x-3) + k \)
• B. \( \log(x+3) + k \)
• C. \( \frac{1}{(x+3)^2} + k \)
• D. \( \frac{x^2}{2} - 3x + k \)

Hint:

When faced with a quadratic in the denominator, factor it and simplify the expression before integrating.

Answer 1

The Correct Option is B

We are given the integral: \[ I = \int \frac{x-3}{x^2-9} \, dx \] First, notice that \( x^2 - 9 = (x-3)(x+3) \). So we can decompose the fraction: \[ \frac{x-3}{x^2 - 9} = \frac{x-3}{(x-3)(x+3)} = \frac{1}{x+3} \] Now, the integral simplifies to: \[ I = \int \frac{1}{x+3} \, dx = \log(x+3) + k \] Thus, the correct answer is \( \log(x+3) + k \).