Compare partial fractions with a split numerator

$\frac{x+1}{(3-x)(3+x)}=\frac A{3-x}+\frac B{3+x}$ gives $x+1=A(3+x)+B(3-x)$. At $x=3$, $A=2/3$; at $x=-3$, $B=-1/3$. Thus the integral is $\int\frac2{3(3-x)}\,dx-\int\frac1{3(3+x)}\,dx=-\frac23\ln|3-x|-\frac13\ln|3+x|+C$.

The derivative of $9-x^2$ is $-2x$. Write the integral as $-\frac12\int\frac{-2(x+1)}{9-x^2}\,dx=-\frac12\int\frac{-2x}{9-x^2}\,dx+\int\frac1{3^2-x^2}\,dx=-\frac12\ln|9-x^2|+\frac16\ln\left|\frac{3+x}{3-x}\right|+C$.

The lecturer note emphasises that the different expressions are equivalent. Use $9-x^2=(3-x)(3+x)$: the second answer becomes $-\frac12\ln|3-x|-\frac12\ln|3+x|+\frac16\ln|3+x|-\frac16\ln|3-x|+C=-\frac23\ln|3-x|-\frac13\ln|3+x|+C$. Both hold on intervals excluding $x=\pm3$.