Put #I=int(3x-4)/{(x-1)(x+3)(x+4)}dx.,# and, notice that the #Dr.# of the rational integrand is made of non-repeated linear factors. In such case, we have to find, #A,B,D in RR# such that,
#(3x-4)/{(x-1)(x+3)(x+4)}=A/(x-1)+B/(x+3)+D/(x+4)............(i)#
#={A(x+3)(x+4)+B(x-1)(x+4)+D(x-1)(x+3)}/{(x-1)(x+3)(x+4)}.#
As the #Drs.# of both sides are same, so must be the #Nrs.# Hence,
#A(x+3)(x+4)+B(x-1)(x+4)+D(x-1)(x+3)=3x-4...........(ii)#
Although #(ii)# holds good #AAx in RR#, but, we chose #x=1, x=-3, and, x=-4# [the reason behind this will soon be clear!] to get,
#A(4)(5)+B*0+D*0=3-4 rArr A=-1/20.#
#A*0+B(-4)(1)+D*0=-9-4 rArr B=13/4.#
#A*0+B*+D(-5)(-1)=-12-4 rArr D=-16/5.# Therefore, by #(i)#
#I=int(-1/20)/(x-1)dx+int(13/4)/(x+3)dx+int(-16/5)/(x+4)dx#
#I=-1/20ln|x-1|+13/4ln|x+3|-16/5ln|x+4|+C.#
Enjoy Maths.!
