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

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Answer 1 · 2017-01-25T14:49:41

The answer is #=-ln(|x|)+ln(|x+1|)+ln(|x-1|)+C#

Let's factorise the denominator

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

So, let's do the decomposition into partial fractions

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

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

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

The denominators are the same, so, we can equalise the numerators

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

Let #x=0#, #=>#, #1=-A#, #=>#, #A=-1#

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

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

Therefore,

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

So, we can do the integration

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

#=-ln(|x|)+ln(|x+1|)+ln(|x-1|)+C#