How do you integrate #int(x+1)/((x-9)(x+8)(x-2))# using partial fractions?

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Answer 1 · 2017-07-10T17:00:43

The answer is #=10/119ln(|x-9|)-7/170ln(|x+8|)-3/70ln(|x-2|)+C#

Let's perform the decomposition into partial fractions

#(x+1)/((x-9)(x+8)(x-2))=A/(x-9)+B/(x+8)+C/(x-2)#

#=(A(x-2)(x+8)+B(x-9)(x-2)+C(x-9)(x+8))/((x-9)(x+8)(x-2))#

The denominator is the same, we compare the numerator.

#x+1=A(x-2)(x+8)+B(x-9)(x-2)+C(x-9)(x+8)#

Let #x=9#, #=>#, #10=17*7A#, #=>#, #A=10/119#

Let #x=-8#, #=>#, #-7=-17*-10B#, #=>#, #B=-7/170#

Let #x=2#, #=>#, #3=-7*10C#, #=>#, #C=-3/70#

Therefore,

#(x+1)/((x-9)(x+8)(x-2))=(10/119)/(x-9)-(7/170)/(x+8)-(3/70)/(x-2)#

#int((x+1)dx)/((x-9)(x+8)(x-2))=int(10/119dx)/(x-9)-int(7/170dx)/(x+8)-int(3/70dx)/(x-2)#

#=10/119ln(|x-9|)-7/170ln(|x+8|)-3/70ln(|x-2|)+C#