# How do you write the partial fraction decomposition of the rational expression (x^2 + 1)/(x(x + 1)(x^2 + 2))?

Dec 14, 2015

$\frac{1}{2 x} - \frac{2}{3 \left(x + 1\right)} + \frac{x + 2}{6 \left({x}^{2} + 2\right)}$

#### Explanation:

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

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

${x}^{2} + 1 = A {x}^{3} + A {x}^{2} + 2 A x + 2 A + B {x}^{3} + 2 B x + C {x}^{3} + C {x}^{2} + D {x}^{2} + D x$

${x}^{2} + 1 = {x}^{3} \left(A + B + C\right) + {x}^{2} \left(A + C + D\right) + x \left(2 A + 2 B + D\right) + 1 \left(2 A\right)$

Thus, $\left\{\begin{matrix}A + B + C = 0 \\ A + C + D = 1 \\ 2 A + 2 B + D = 0 \\ 2 A = 1\end{matrix}\right.$

Solve this to see that $\left\{\begin{matrix}A = \frac{1}{2} \\ B = - \frac{2}{3} \\ C = \frac{1}{6} \\ D = \frac{1}{3}\end{matrix}\right.$

Therefore,

$\frac{{x}^{2} + 1}{x \left(x + 1\right) \left({x}^{2} + 2\right)} = \frac{1}{2 x} - \frac{2}{3 \left(x + 1\right)} + \frac{x + 2}{6 \left({x}^{2} + 2\right)}$