Помогите пожалуйста ,задание 4.

Решите задачу:

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