How do you integrate #(x^2-1)/(x^2-16)# using partial fractions?

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Answer 1 · 2016-12-02T06:41:31

The answer is #=x-15/8ln(∣x+4∣)+15/8ln(∣x-4∣)+C#

Firstly

#(x^2-1)/(x^2-16)=1+15/(x^2-16)#

#x^2-16=(x+4)(x-1)#

Now we can do the decomposition in partial fractions

#15/(x^2-16)=15/((x+4)(x-1))=A/(x+4)+B/(x-4)#

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

Therefore,

#15=A(x-4)+B(x+4)#

Let #x=4#, #=>#, #15=8B#

#B=15/8#

Let #x=-4#, #=>#, #15=-8A#

#A=-15/8#

#(x^2-1)/(x^2-16)=1-15/(8(x+4))+15/(8(x-4)#

#int((x^2-1)dx)/(x^2-16)=int1dx-int(15dx)/(8(x+4))+int(15dx)/(8(x-4)#

#=x-15/8ln(∣x+4∣)+15/8ln(∣x-4∣)+C#