Question

Evaluate: \(\displaystyle \int \frac{dx}{x(x+2)} = \)

• A. \( \frac{1}{2} \log \left| \frac{x}{x+2} \right| + c \)
• B. \( \frac{1}{2} \log \left| \frac{x+2}{x} \right| + c \)
• C. \( \log|x| - \log|x+2| + c \)
• D. \( \log|x| + \log|x+2| + c \)

Hint:

For integrals involving rational functions with quadratic denominators, try partial fraction decomposition: \[ \frac{1}{(x+a)(x+b)} = \frac{A}{x+a} + \frac{B}{x+b} \] Then integrate term by term.

Answer 1

The Correct Option is B

Step 1: Use partial fractions: \[ \frac{1}{x(x+2)} = \frac{A}{x} + \frac{B}{x+2} \] Multiply both sides by \( x(x+2) \): \[ 1 = A(x+2) + Bx = (A+B)x + 2A \] Step 2: Equate coefficients: \[ A + B = 0, \quad 2A = 1 \Rightarrow A = \frac{1}{2}, \quad B = -\frac{1}{2} \] Step 3: Rewrite integral: \[ \int \frac{dx}{x(x+2)} = \int \left( \frac{1/2}{x} - \frac{1/2}{x+2} \right) dx = \frac{1}{2} \int \frac{dx}{x} - \frac{1}{2} \int \frac{dx}{x+2} \] Step 4: Integrate: \[ = \frac{1}{2} \log|x| - \frac{1}{2} \log|x+2| + c = \frac{1}{2} \log \left| \frac{x}{x+2} \right| + c \] Step 5: Or equivalently, \[ = \frac{1}{2} \log \left| \frac{x}{x+2} \right| + c = -\frac{1}{2} \log \left| \frac{x+2}{x} \right| + c \] Depending on sign convention.