Найти производную четвертого порядка y=(x^2 +3)ln(x-1)

Решите задачу:

$y=(x^2+3)ln(x-1)\\\\y'=2xln(x-1)+\frac{x^2+3}{x-1}\\\\y''=2ln(x-1)+\frac{2x}{x-1}+\frac{2x(x-1)-(x^2+3)}{(x-1)^2}=\\\\=2ln(x-1)+\frac{2x}{x-1}+\frac{x^2-2x-3}{(x-1)^2}\\\\y'''=\frac{2}{x-1}+\frac{2(x-1)-2x}{(x-1)^2}+\frac{(2x-2)-(x^2-2x-3)\cdot 2(x-1)}{(x-1)^4}=\\\\=\frac{2}{x-1}+\frac{-1}{(x-1)^2}+\frac{2(x-1)(1-x^2+2x+3)}{(x-1)^4}=\frac{2}{x-1}-\frac{2}{(x-1)^2}+\frac{2(4+2x-x^2)}{(x-1)^3}\\\\y^{(4)}=\frac{-2}{(x-1)^2}-\frac{2\cdot 2(x-1)}{(x-1)^4}+\frac{2(2-2x)(x-1)^3-2(4+2x-x^2)3(x-1)^2}{(x-1)^6}$

$=-\frac{2}{(x-1)^2}-\frac{4}{(x-1)^3}-\frac{2\cdot (14+2x-x^2)}{(x-1)^4}$