Помогите сделать последний пример

$(\frac{x+1}{3x-3}+\frac{6}{3x^{2}-3}-\frac{x+3}{3*(x+3)})*\frac{3x^{2}-3}{4}=\frac{x+4}{2}$

1) $\frac{x+1}{3*(x-1)}+\frac{6}{3*(x-1)(x+1)}-\frac{x+3}{3*(x+3)}=\frac{(x+1)^{2}(x+3)+6(x+3)-(x+3)(x^{2}-1)}{3*(x-1)*(x+1)(x+3)}=$ $\frac{(x^{2}+2x+1)(x+3)+6x+18-(x^{3}-x+3x^{2}-3)}{3*(x-1)*(x+1)(x+3)}=$ $\frac{x^{3}+2x^{2}+x+3x^{2}+6x+3+6x+18-x^{3}+x-3x^{2}+3}{3*(x-1)*(x+1)(x+3)}=\frac{2x^{2}+14x+24}{3*(x-1)*(x+1)(x+3)}$ $=\frac{2(x^{2}+7x+12)}{3*(x-1)*(x+1)(x+3)}=$ $\frac{2(x+3)(x+4)}{3*(x-1)*(x+1)(x+3)}=\frac{2(x+4)}{3*(x-1)*(x+1)}$

2) $\frac{2(x+4)}{3*(x-1)*(x+1)}*\frac{3(x-1)(x+1)}{4}=\frac{x+4}{2}$