## University Calculus: Early Transcendentals (3rd Edition)

$\frac{1}{x+1}+\frac{3}{(x+1)^{2}}$
The form of the partial fraction decomposition is $\frac{x+4}{(x+1)^{2}}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}$ This gives $A(x+1)+B= x+4$ Equating the coefficients of x and the constant term, we get A=1, A+B=4 or B=3. Thus we can write $\frac{x+4}{(x+1)^{2}}= \frac{1}{x+1}+\frac{3}{(x+1)^{2}}$