Consider the indefinite integral

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{3x^3+3x^2-40x-48}{x^2-16},dx$

Then the integrand decomposes into the form

$ax+b+\frac{c}{x-4}+\frac{d}{x+4}$

where

$a=$

$b=$

$c=$

$d=$

Integrating term $by$ term, $we$ obtain that

${\displaystyle \underset{+C}{\overset{\phantom{}}{}}}\frac{3x^3+3x^2-40x-48}{x^2-16},dx=$

$+C$