Find $\frac{\mathrm d^{100}}{\mathrm d x^{100}}\frac{x^2+1}{x^3-x}=$?

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

$$\frac{d^n(y^{-1})}{dy^n} = \frac{(-1)^n n!}{y^{n+1}}$$

Hence, the $n^{th}$ derivative is $$\frac{(-1)^{n+1} n!}{x^{n+1}} + \frac{(-1)^n n!}{(x-1)^{n+1}} + \frac{(-1)^n n!}{(x+1)^{n+1}} = (-1)^n n! \times \left( \frac{-1}{x^{n+1}} + \frac{1}{(x-1)^{n+1}} + \frac{1}{(x+1)^{n+1}}\right)$$

Similarly, $$\frac{1}{x^2-3x+2} = \frac{1}{x-2} - \frac{1}{x-1}$$

Hence, the $n^{th}$ derivative is $$\frac{(-1)^n n!}{(x-2)^{n+1}} - \frac{(-1)^n n!}{(x-1)^{n+1}} = (-1)^n n! \times \left( \frac{1}{(x-2)^{n+1}} - \frac{1}{(x-1)^{n+1}}\right)$$

HINT: Try using partial fraction decomposition.