$\frac{1}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}….(1)$

Which implies $A(x+2)+Bx=1$, putting $x+2=0$

Therefore, $x=-2$,

And $B=-0.5$

From equation (1)

$\frac{1}{x(x+2)}=\frac{1}{2}\times \frac{1}{x}-\frac{1}{2}\times \frac{1}{x+2}$

$\int \frac{1}{x(x+2)}dx=\frac{1}{2}\int \frac{1}{x}dx-\frac{1}{2}\int \frac{1}{x+2}dx$

$\frac{1}{2}\log \left| x \right|-\frac{1}{2}\log \left| x+2 \right|+c$

$\frac{1}{2}[\log \left| x \right|-\log \left| x+2 \right|]+c$

$\frac{1}{2}\log \left| \frac{x}{x+2} \right|+c$