Question

One of the partial fractions of \[ \frac{2x^2+x-3}{(x^2+2)(3x-1)} \] is

• A. \[ \frac{22}{19(3x-1)} \]
• B. \[ \frac{20x-13}{19(x^2+2)} \]
• C. \[ \frac{20x+13}{19(x^2+2)} \]
• D. \[ \frac{22}{3x-1} \] \bigskip

Hint:

For a denominator containing a quadratic factor and a linear factor, \[ \frac{P(x)}{(x^2+a)(bx+c)} = \frac{Ax+B}{x^2+a} + \frac{C}{bx+c}. \] After clearing denominators, compare coefficients of equal powers of \(x\) to find the unknown constants.

Answer 1

The Correct Option is C

Concept: Decompose the rational function into partial fractions by assuming unknown constants and comparing coefficients.

Step 1: Assume \(\frac{2x^2+x-3}{(x^2+2)(3x-1)}=\frac{Ax+B}{x^2+2}+\frac{C}{3x-1}\).

Step 2: Multiplying throughout by \((x^2+2)(3x-1)\) and comparing coefficients gives \(3A+C=2,\;-A+3B=1,\;-B+2C=-3\). Solving, \(A=\frac{20}{19},\;B=\frac{13}{19},\;C=-\frac{22}{19}\).

Step 3: Therefore, \(\displaystyle \frac{2x^2+x-3}{(x^2+2)(3x-1)}=\frac{20x+13}{19(x^2+2)}-\frac{22}{19(3x-1)}\), so the required partial fraction is \(\boxed{\frac{20x+13}{19(x^2+2)}}\).