Помогите решить неопределенный интеграл∫ dx/(3x^2+2x-1)

Решите задачу:

$\int\frac{dx}{3x^2+2x-1}=\int\frac{dx}{3(x+1)(x-\frac{1}{3})}=\int\frac{dx}{(x+1)(3x-1)}=\\\\=[\frac{1}{(x+1)(3x-1)}=\frac{A}{x+1}+\frac{B}{3x-1}=\frac{A(3x-1)+B(x+1)}{(x+1)(3x-1)}\Rightarrow \\\Rightarrow1=A(3x-1)+B(x+1)\\x=-1:\ \ \ \ 1=A(3*(-1)-1)+B*0\rightarrow A=-\frac{1}{4}\\x=\frac{1}{3}:\ \ \ \ \ \ 1=A*0+B(\frac{1}{3}+1)\rightarrow B=\frac{3}{4}\\\frac{1}{(x+1)(3x-1)}=\frac{-\frac{1}{4}}{x+1}+\frac{\frac{3}{4}}{3x-1}=\frac{3}{4(3x-1)}-\frac{1}{4(x+1)}]=$

$\\\\\int(\frac{3}{4*3(x-\frac{1}{3})}-\frac{1}{4(x+1)})dx=\frac{1}{4}\int\frac{1}{x-\frac{1}{3}}-\frac{1}{4}\int\frac{1}{x+1}=\\=\frac{1}{4}ln|x-\frac{1}{3}|-\frac{1}{4}ln|x+1|+C$