Which of the following is a partial fraction of $\frac{-x^2+6x+13}{(3x+5)\left(x^2+4x+4\right)}=$

Given, $\frac{-x^2+6x+13}{(3x+5)\left(x^2+4x+4\right)}$

$=\frac{-x^2+6x+13}{(3x+5){\left(x+2\right)}^2}$

Using Partial fraction rules, we get

$\frac{-x^2+6x+13}{(3x+5){\left(x+2\right)}^2}=\frac{A}{3x+5}+\frac{B}{x+2}+\frac{C}{{\left(x+2\right)}^2}$

$\Rightarrow -x^2+6x+13=A{\left(x+2\right)}^2+B\left(x+2\right)\left(3x+5\right)+C\left(3x+5\right)$

Putting $x=-2$, we get $C=3$

Putting $x=\frac{-5}{3}$, we get $A=2$

and on comparing coefficients of $x^2$, we get

$B=-1$

i.e. $\frac{-x^2+6x+13}{(3x+5){\left(x+2\right)}^2}=\frac{2}{3x+5}-\frac{1}{x+2}+\frac{3}{{\left(x+2\right)}^2}$