How do you integrate #int (-2x-3)/(x^2-x) # using partial fractions?

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Answer 1 · 2016-11-17T16:59:19

The answer is #=3lnx-5ln(x-1)+C#

Let's do the decomposition in partial fractions

The denominator #=x^2-x=x(x-1)#

#(-2x-3)/(x(x-1))=A/x+B/(x-1)#

#=(A(x-1)+Bx)/(x(x-1))#

Therefore,
#-2x-3=A(x-1)+Bx#

let #x=1#, #=># #-5=B#

let #x=0#, #=># #-3=-A# #=># #A=3#

So, #(-2x-3)/(x(x-1))=3/x-5/(x-1)#

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

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

as #intdx/x=lnx+c#