Помогите пожалуйста с интегралами

Решите задачу:

$\int\frac{dx}{x^2-9}=\int\frac{dx}{(x-3)(x+3)}=[\frac{1}{(x-3)(x+3)}=\frac{A}{x-3}+\frac{B}{x+3}=\frac{A(x+3)+B(x-3)}{(x-3)(x+3)}\Rightarrow\\\Rightarrow1=A(x+3)+B(x-3)\\x=-3:\ \ 1=A(-3+3)+B(-3-3)\rightarrow B=-\frac{1}{6}\\x=3:\ \ \ \ 1=A(3+3)+B(3-3)\rightarrow A=\frac{1}{6}\\\frac{1}{(x-3)(x+3)}=\frac{1}{6(x-3)}-\frac{1}{6(x+3)}]=\int(\frac{1}{6(x-3)}-\frac{1}{6(x+3)})dx=[d(x-3)=dx\\d(x+3)=dx]=$
$=\frac{1}{6}\int\frac{d(x-3)}{x-3}-\frac{1}{6}\int\frac{d(x+3)}{x+3}=\frac{1}{6}(ln|x-3|-ln|x+3|)+C$

$\int\frac{dx}{xln^2x}=[d(lnx)=\frac{1}{x}dx\rightarrow dx=x*d(lnx)]=\int\frac{xd(lnx)}{xln^2x}=\\=\int\frac{d(lnx)}{ln^2x}=\frac{(lnx)^{-2+1}}{-2+1}+C=-\frac{1}{lnx}+C$

$\int\frac{xdx}{(x^2-3)^4}=[d(x^2-3)=2xdx\rightarrow dx=\frac{d(x^2-3)}{2x}]=\int\frac{x}{(x^2-3)^4}*\frac{d(x^2-3)}{2x}=\\=\frac{1}{2}\int\frac{d(x^2-3)}{(x^2-3)^4}=\frac{1}{2}*\frac{(x^2-3)^{-4+1}}{-4+1}+C=-\frac{1}{6}*\frac{1}{(x^2-3)^3}+C$