Question

If \[ \frac{x}{(x-1)^2(x+2)} = \frac{A}{(x-1)^2} + \frac{2}{9(x-1)} + \frac{B}{x+2}, \] then \(A+B=\)

• A. \(\frac{1}{3}\)
• B. \(\frac{1}{9}\)
• C. \(-\frac{1}{3}\)
• D. \(\frac{2}{3}\)

Hint:

In partial fractions, substitute the roots of denominator factors to quickly find unknown constants.

Answer 1

The Correct Option is B

Concept: Partial fractions are used to split a rational expression into simpler fractions. Here the denominator has a repeated factor \((x-1)^2\), so both \((x-1)^2\) and \((x-1)\) terms are involved.

Step 1: Write the given expression. \[ \frac{x}{(x-1)^2(x+2)} = \frac{A}{(x-1)^2} + \frac{2}{9(x-1)} + \frac{B}{x+2} \]

Step 2: Multiply both sides by \((x-1)^2(x+2)\). \[ x=A(x+2)+\frac{2}{9}(x-1)(x+2)+B(x-1)^2 \]

Step 3: Put \(x=1\) to find \(A\). \[ 1=A(1+2) \] \[ 1=3A \] \[ A=\frac{1}{3} \]

Step 4: Put \(x=-2\) to find \(B\). \[ -2=B(-2-1)^2 \] \[ -2=9B \] \[ B=-\frac{2}{9} \]

Step 5: Find \(A+B\). \[ A+B=\frac{1}{3}-\frac{2}{9} \] \[ A+B=\frac{3}{9}-\frac{2}{9} \] \[ A+B=\frac{1}{9} \] Therefore, \[ \boxed{\frac{1}{9}} \]