How do you express (x^2-2x-1) / ((x-1)^2 (x^2+1)) in partial fractions?

Nov 15, 2017

The answer is $= - \frac{1}{x - 1} ^ 2 + \frac{1}{x - 1} - \frac{\left(x - 1\right)}{{x}^{2} + 1}$

Explanation:

Perform the decomposition into partial fractions

$\frac{{x}^{2} - 2 x - 1}{{\left(x - 1\right)}^{2} \left({x}^{2} + 1\right)} = \frac{A}{x - 1} ^ 2 + \frac{B}{x - 1} + \frac{C x + D}{{x}^{2} + 1}$

$= \frac{A \left({x}^{2} + 1\right) + B \left(x - 1\right) \left({x}^{2} + 1\right) + \left(C x + D\right) {\left(x - 1\right)}^{2}}{{\left(x - 1\right)}^{2} \left({x}^{2} + 1\right)}$

The denominators are the same, compare the numerators

$\left({x}^{2} - 2 x - 1\right) = A \left({x}^{2} + 1\right) + B \left(x - 1\right) \left({x}^{2} + 1\right) + \left(C x + D\right) {\left(x - 1\right)}^{2}$

Let $x = 1$

$- 2 = 2 A$, $A = - 1$

Coefficients of ${x}^{2}$

$1 = A - B - 2 C + D$

Coefficients of $x$

$- 2 = B + C - 2 D$

and

$- 1 = A - B + D$, $\implies$, $- B + D = 0$, $B = D$

$1 = - 1 - B - 2 C + B$, $\implies$, $2 C = - 2$, $\implies$, $C = - 1$

$- 2 = B - 1 - 2 B$, $\implies$, $B = 1$

$\implies$, $D = 1$

Finally,

$\frac{{x}^{2} - 2 x - 1}{{\left(x - 1\right)}^{2} \left({x}^{2} + 1\right)} = - \frac{1}{x - 1} ^ 2 + \frac{1}{x - 1} - \frac{\left(x - 1\right)}{{x}^{2} + 1}$