How do you integrate #int(x+1)/((x+5)(x-1)(x-4))# using partial fractions?

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Answer 1 · 2017-09-01T19:00:02

The integral is #=-2/27ln(|x+5|)-1/9ln(|x-1|)+5/27ln(|x-4|)+C#

Let's perform the decomposition into partial fractions

#(x+1)/((x+5)(x-1)(x-4))=A/(x+5)+B/(x-1)+C/(x-4)#

#=(A(x-1)(x-4)+B(x+5)(x-4)+C(x+5)(x-1))/((x+5)(x-1)(x-4))#

The denominators are the same, we compare the numerators

#x+1=A(x-1)(x-4)+B(x+5)(x-4)+C(x+5)(x-1)#

Let #x=-5#, #=>#, #-4=A*(-6)(-9)#, #=>#, #A=-2/27#

Let #x=1#, #=>#, #2=B*(6)(-3)#, #=>#, #B=-1/9#

Let #x=4#, #=>#, #5=C*(9)(3)#, #=>#, #C=5/27#

Therefore,

#(x+1)/((x+5)(x-1)(x-4))=(-2/27)/(x+5)+(-1/9)/(x-1)+(5/27)/(x-4)#

#int((x+1)dx)/((x+5)(x-1)(x-4))=int(-2/27dx)/(x+5)+int(-1/9dx)/(x-1)+int(5/27dx)/(x-4)#

#=-2/27ln(|x+5|)-1/9ln(|x-1|)+5/27ln(|x-4|)+C#