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Question: The partial fractions of \(\frac{3x - 1}{(1 - x + x^{2})(2 + x)}\) are...

The partial fractions of 3x−1(1−x+x2)(2+x)\frac{3x - 1}{(1 - x + x^{2})(2 + x)} are

A

x(x2−x+1)\frac{x}{(x^{2} - x + 1)}+1x+2\frac{1}{x + 2}

B

1x2−x+1+xx+2\frac{1}{x^{2} - x + 1} + \frac{x}{x + 2}

C

xx2−x+1−1x+2\frac{x}{x^{2} - x + 1} - \frac{1}{x + 2}

D

−1x2−x+1+xx+2\frac{- 1}{x^{2} - x + 1} + \frac{x}{x + 2}

Answer

xx2−x+1−1x+2\frac{x}{x^{2} - x + 1} - \frac{1}{x + 2}

Explanation

Solution

3x−1(1−x+x2)(2+x)=Ax+Bx2−x+1+Cx+2\frac{3x - 1}{(1 - x + x^{2})(2 + x)} = \frac{Ax + B}{x^{2} - x + 1} + \frac{C}{x + 2}

⇒ (3x−1)=(Ax+B)(x+2)+C(x2−x+1)(3x - 1) = (Ax + B)(x + 2) + C(x^{2} - x + 1) Comparing the

coefficient of like terms, we get A+C=0A + C = 0, 2A+B−C=32A + B - C = 3, 2B+C=−12B + C = - 1 ⇒ A=1A = 1, B=0B = 0, C=−1C = - 1

∴\therefore 3x−1(1−x+x2)(2+x)=xx2−x+1−1x+2\frac{3x - 1}{(1 - x + x^{2})(2 + x)} = \frac{x}{x^{2} - x + 1} - \frac{1}{x + 2}.