$\begin{aligned}
& \text {Given, } \frac{2 x^3+x^2-5}{x^4-25}=\frac{A x+B}{x^2-5}+\frac{C x+1}{x^2+5} \\
& \therefore 2 x^3+x^2-5=(A x+B)\left(2 x^2+5\right) \\
& +(C x+1)\left(x^2-5\right) \\
& \Rightarrow 2 x^3+x^2-5=A x^3+5 A x+B x^2+5 B \\
& +C x^3-5 C x+x^2-5 \\
& \Rightarrow 2 x^3+x^2-5=x^3(A+C)+x^2(B+1) \\
& +x(5 A-5 C)+5 B-5
\end{aligned}$
On equating the coefficients of $x^3, x^2, x$ and constant, we get
$\begin{aligned}
& A+C=2, B+1=1,5 A-5 C=0 \\
& \text {and } \\
& 5 B-5=-5 \\
& B+1=1 \Rightarrow B=0 \\
& 5 A-5 C=0 \Rightarrow A=C \\
& A+C=2 \Rightarrow C+C=2 \\
& \Rightarrow \quad 2 C=2 \Rightarrow C=1
\end{aligned}$
Now,
and
$\begin{aligned} & \therefore & A & =C=1 \\ & \therefore & (A, B, C) & =(1,0,1)\end{aligned}$