\(\displaystyle \frac{A}{x-3}+\frac{B}{(x-3)^{2}}=\frac{3x}{(x - 3)^2}\)
Multiply both sides of the decomposition by the LCD, \(\displaystyle (x-3)^{2}\)
\(\displaystyle (x-3)^{2}\cdot \frac{A}{x-3}+\frac{B}{(x-3)^{2}}\cdot (x-3)^{2}=\frac{3x}{(x - 3)^2} \cdot (x - 3)^2\)
\(\displaystyle A(x-3)+B=3x\)
\(\displaystyle Ax-3A+B=3x\)
Equate coefficients:
\(\displaystyle A=3\)
\(\displaystyle -3A+B=0\)
Solve for A and B.
It is obvious that A=3 and B=9
\(\displaystyle \frac{3}{x-3}+\frac{9}{(x-3)^{2}}\)
