How do you integrate #int(x+1)/((x+5)(x+2)(x-5))# using partial fractions?

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Answer 1 · 2017-01-10T14:28:28

The answer is #=-2/15(∣x+5∣)+1/21ln(∣x+2∣)+3/35ln(∣x-5∣)+C#

Perform the decomposition into partial fractions

#(x+1)/((x+5)(x+2)(x-5))=A/(x+5)+B/(x+2)+C/(x-5)#

#=(A(x+2)(x-5)+B(x+5)(x-5)+C(x+5)(x+2))/((x+5)(x+2)(x-5))#

Therefore, by equating the numerators

#(x+1)=(A(x+2)(x-5)+B(x+5)(x-5)+C(x+5)(x+2))#

Let #x=-5#, #=>#, #-4=30A#, #=>#, #A=-2/15#

Let #x=-2#, #=>#, #-1=-21B#, #=>#, #B=1/21#

Let #x=5#, #=>#, #6=70C#, #=>#, #C=3/35#

so,

#(x+1)/((x+5)(x+2)(x-5))=(-2/15)/(x+5)+(1/21)/(x+2)+(3/35)/(x-5)#

Now, perform the integration

#int((x+1)dx)/((x+5)(x+2)(x-5))=-2/15intdx/(x+5)+1/21intdx/(x+2)+3/35intdx/(x-5)#

#=-2/15(∣x+5∣)+1/21ln(∣x+2∣)+3/35ln(∣x-5∣)+C#