Помогите: интеграл x^3dx/(x-1)^2(x+3)

Решите задачу:

$\int \frac{x^3\, dx}{(x-1)^2(x+3)}=\int \frac{x^3\, dx}{(x^2-2x+1)(x+3)}=\int \frac{x^3}{x^3+x^2-5x+1}dx=\\\\=\int (1+\frac{-x^2+5x-1}{x^3+x^2-5x+1})dx=\int dx-\int \frac{x^2-5x+1}{(x-1)^2(x+3)}dx;\\\\\frac{x^2-5x+1}{(x-1)^2(x+3)}=\frac{A}{(x-1)^2}+\frac{B}{x-1}+\frac{C}{x+3}\quad \to \\\\x^2-5x+1=A(x+3)+B(x-1)(x+3)+C(x-1)^2\; \; \to$

$x=1:\; \; A=\frac{1-5+1}{1+3}=-\frac{3}{4}\\\\x=-3:\; \; C=\frac{9+15+1}{16}=\frac{25}{16}\\\\x^2|\; \; 1=B+C\; \; \; \rightarrow \; \; \; B=C-1=\frac{25}{16}-1=\frac{9}{16}$

$\int \frac{x^3\, dx}{(x-1)^2(x+3)}=x+\frac{3}{4}\int \frac{dx}{(x-1)^2}-\frac{9}{16}\int \frac{dx}{x-1}-\frac{25}{16}\int \frac{dx}{x+3}=\\\\=x+\frac{3}{4}\cdot \frac{(x-1)^3}{3}-\frac{25}{16}\cdot ln|x+3|+C$