## Thomas' Calculus 13th Edition

$\frac{1}{x+1}+\frac{3}{(x+1)^{2}}$
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, $\frac{x+4}{(x+1)^{2}}=\frac{1}{x+1}+\frac{3}{(x+1)^{2}}$