How do I find the integral #int1/((w-4)(w+1))dw# ?

1 Answer
Jul 28, 2014

#=1/5ln((w-4)/(w+1))+c#, where #c# is a constant

Explanation :

This type of question usually solve by using Partial Fractions,

#1/((w-4)(w+1))#, it can be written as

#1/((w-4)(w+1))=A/(w-4)+B/(w+1)#

multiplying by #(w-4)(w+1)# on both sides, we get

#1=A(w+1)+B(w-4)#

#1=(A+B)w+(A-4B)#

Now comparing coefficient of #w# and constants both sides, we get

#A+B=0# #=># #A=-B# ...........#(i)#
#A-4B=1# ..............#(ii)#

Substituting value of #A# from #(i)# to #(ii)#, we get

#-5B=1# #=># #B=-1/5#

from #B#, we can easily calculate #A#, which will be #1/5#

Now,

#1/((w-4)(w+1))=1/(5(w-4))-1/(5(w+1))#

Integrating both side with respect to #w#,

#int1/((w-4)(w+1))dw=int1/(5(w-4))dw-int1/(5(w+1))dw#

#=1/5(ln(w-4)-ln(w+1))+c#, where #c# is a constant

#=1/5ln((w-4)/(w+1))+c#, where #c# is a constant