Consider the indefinite integral ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{6x^{3}3x^{2}2x-3}{x^2-1}dx$

Then the integrand decomposes into the form

$axb\frac{c}{x-1}\frac{d}{x1}$

where

$a=$

$b=$

$c=$

$d=$

Integrating term by term, we obtain that

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{6x^{3}3x^{2}2x-3}{x^2-1}dx=$

$C$