If (\frac{(x - 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{Bx +... | MATH - PARTIAL FRACTIONS | SolveeIt

If $\frac{(x - 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1}$ , then

$A = 1,B = 0,C = 2$

$A = 1,B = 0,C = - 2$

$A = - 1,B = 0,C = - 2$

None of these

Answer

$A = 1,B = 0,C = - 2$

Explanation

$A(x^{2} + 1) + x(Bx + C) = (x - 1)^{2}$

For $x = i, - B + Ci = - 2i$ $\Rightarrow$ $B = 0,C = - 2$

Equating coefficient of $x^{2}$ ,

$A + B = 1 \Rightarrow A = 1 - B = 1 - 0 = 1$ ;

$\therefore A = 1,B = 0,C = - 2$ .

Question: If \(\frac{(x - 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1}\), then...