#### Answer

$\frac{1}{x+1}+\frac{3}{(x+1)^{2}}$

#### Work Step by Step

If the form of the rational function is $\frac{px+q}{(x-a)^{2}}$, then the form of the partial fraction is $\frac{A}{x-a}+\frac{B}{(x-b)^{2}}$.
Therefore,
$\frac{x+4}{(x+1)^{2}}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}$
A and B are to be determined.
x+4= A(x+1)+B
Equating the coefficients of x and the constant term, we get
A+B=4 and A=1. Then, B=4-1=3.
Therefore,
$\frac{x+4}{(x+1)^{2}}=\frac{1}{x+1}+\frac{3}{(x+1)^{2}}$