How do you find #int (11-2x)/(x^2 + x - 2) dx# using partial fractions?

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Answer 1 · 2016-10-22T05:49:51

#int((11-2x)dx)/((x-1)(x+2))=3ln(x-1)-5ln(x+2)+C#

First factorise the denominator
#x^2+x-2=(x-1)(x+2)#

Then we look for the partial fraction

#(11-2x)/((x-1)(x+2))=A/(x-1)+B/(x+2)#

#=(A(x+2) +B(x-1))/((x-1)(x+2))#

So #11-2x=A(x+2) +B(x-1)#

If #x=1# #=># #11-2=3A+0#
so #A=9/3=3#

If #x=-2# #=># #11+4=0-3B#
so #B=15/-3=-5#

#(11-2x)/((x-1)(x+2))=3/(x-1)-5/(x+2)#

Then we can integrate

#int((11-2x)dx)/((x-1)(x+2))=int(3dx)/(x-1)-int(5dx)/(x+2)#

#=3ln(x-1)-5ln(x+2)+C#

Answer 2 · 2016-10-23T02:40:23

Oct 23, 2016

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