Question:
One of the partial fractions of
\[
\frac{2x^2+x-3}{(x^2+2)(3x-1)}
\]
is
One of the partial fractions of
\[
\frac{2x^2+x-3}{(x^2+2)(3x-1)}
\]
is
Show 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.
Updated On: Jul 9, 2026
- \[ \frac{22}{19(3x-1)} \]
- \[ \frac{20x-13}{19(x^2+2)} \]
- \[ \frac{20x+13}{19(x^2+2)} \]
- \[ \frac{22}{3x-1} \] \bigskip
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
When a rational function has a quadratic factor and a linear factor in the denominator, its partial fraction decomposition is of the form
\[
\frac{P(x)}{(x^2+2)(3x-1)}
=
\frac{Ax+B}{x^2+2}
+
\frac{C}{3x-1}.
\]
We determine the constants \(A\), \(B\), and \(C\) by comparing coefficients.
Step 1: Assume the required partial fraction form. Let \[ \frac{2x^2+x-3}{(x^2+2)(3x-1)} = \frac{Ax+B}{x^2+2} + \frac{C}{3x-1}. \] Multiplying throughout by \[ (x^2+2)(3x-1), \] we get \[ 2x^2+x-3 = (Ax+B)(3x-1)+C(x^2+2). \]
Step 2: Expand and compare coefficients. Expanding, \[ (Ax+B)(3x-1) = 3Ax^2-Ax+3Bx-B. \] Hence, \[ 2x^2+x-3 = (3A+C)x^2+(-A+3B)x+(-B+2C). \] Comparing coefficients of like powers of \(x\), \[ 3A+C=2, \] \[ -A+3B=1, \] \[ -B+2C=-3. \]
Step 3: Solve for \(A\), \(B\), and \(C\). From \[ 3A+C=2, \] \[ C=2-3A. \] Substituting into \[ -B+2C=-3, \] \[ -B+2(2-3A)=-3, \] \[ B=7-6A. \] Using \[ -A+3B=1, \] \[ -A+3(7-6A)=1. \] \[ -A+21-18A=1. \] \[ -19A=-20. \] \[ A=\frac{20}{19}. \] Therefore, \[ B=7-\frac{120}{19} =\frac{13}{19}, \] and \[ C=2-\frac{60}{19} =-\frac{22}{19}. \]
Step 4: Write the partial fraction decomposition. Substituting the values, \[ \frac{2x^2+x-3}{(x^2+2)(3x-1)} = \frac{\frac{20}{19}x+\frac{13}{19}}{x^2+2} -\frac{22}{19(3x-1)}. \] \[ = \frac{20x+13}{19(x^2+2)} -\frac{22}{19(3x-1)}. \] Thus, one of the partial fractions is \[ \frac{20x+13}{19(x^2+2)}. \]
Step 5: Write the final answer. \[ \boxed{\frac{20x+13}{19(x^2+2)}} \]
Step 1: Assume the required partial fraction form. Let \[ \frac{2x^2+x-3}{(x^2+2)(3x-1)} = \frac{Ax+B}{x^2+2} + \frac{C}{3x-1}. \] Multiplying throughout by \[ (x^2+2)(3x-1), \] we get \[ 2x^2+x-3 = (Ax+B)(3x-1)+C(x^2+2). \]
Step 2: Expand and compare coefficients. Expanding, \[ (Ax+B)(3x-1) = 3Ax^2-Ax+3Bx-B. \] Hence, \[ 2x^2+x-3 = (3A+C)x^2+(-A+3B)x+(-B+2C). \] Comparing coefficients of like powers of \(x\), \[ 3A+C=2, \] \[ -A+3B=1, \] \[ -B+2C=-3. \]
Step 3: Solve for \(A\), \(B\), and \(C\). From \[ 3A+C=2, \] \[ C=2-3A. \] Substituting into \[ -B+2C=-3, \] \[ -B+2(2-3A)=-3, \] \[ B=7-6A. \] Using \[ -A+3B=1, \] \[ -A+3(7-6A)=1. \] \[ -A+21-18A=1. \] \[ -19A=-20. \] \[ A=\frac{20}{19}. \] Therefore, \[ B=7-\frac{120}{19} =\frac{13}{19}, \] and \[ C=2-\frac{60}{19} =-\frac{22}{19}. \]
Step 4: Write the partial fraction decomposition. Substituting the values, \[ \frac{2x^2+x-3}{(x^2+2)(3x-1)} = \frac{\frac{20}{19}x+\frac{13}{19}}{x^2+2} -\frac{22}{19(3x-1)}. \] \[ = \frac{20x+13}{19(x^2+2)} -\frac{22}{19(3x-1)}. \] Thus, one of the partial fractions is \[ \frac{20x+13}{19(x^2+2)}. \]
Step 5: Write the final answer. \[ \boxed{\frac{20x+13}{19(x^2+2)}} \]
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