If  $\frac{x+1}{(x-5)(x+5)}=\frac{1}{k}\left[\frac{A}{x-5}+\frac{B}{x+5}\right]$ then  $k=$

$\frac{x+1}{(x-5)(x+5)}=\frac{1}{k}\left(\frac{A}{x-5}+\frac{B}{x+5}\right)$

$\frac{x+1}{(x-5)(x+5)}=\frac{1}{k}\left(\frac{A(x+5)+B(x-5)}{(x-5)(x+5)}\right)$

$k(x+1)=A(x+5)+B(x-5)$

$putx=5\Rightarrow 6k=10A\Rightarrow A=\frac{6k}{10}=\frac{3k}{5}$

$putx=-5\Rightarrow -4k=-10B\Rightarrow B=\frac{2k}{5}$

$\frac{x+1}{(x-5)(x+5)}=\frac{1}{k}\left(\frac{\frac{3k}{5}}{x-5}+\frac{\frac{2k}{5}}{x+5}\right)$

$\frac{x+1}{(x-5)(x+5)}=\frac{1}{5}\left(\frac{3}{x-5}+\frac{2}{x+5}\right)$

$\Rightarrow k=5$