How do you use partial fractions to find the integral #int 1/(x^2-1)dx#?

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Answer 1 · 2016-12-09T06:28:11

The answer is #==-1/2ln(∣x+1∣)+1/2ln(∣x-1∣)+C#

We use #a^2-b^2=(a+b)(a-b)#

The denominator is

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

So, the decompostion into partial fractions is

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

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

Therefore,

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

Let #x=1#, #=>#, #1=2B#

Let #x=-1#, #=>#,#1=-2A#

so,

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

#int(1dx)/(x^2-1)=int(-1dx)/(2(x+1))+int(1dx)/(2(x-1))#

#=-1/2ln(∣x+1∣)+1/2ln(∣x-1∣)+C#