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

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Answer 1 · 2016-12-11T07:01:40

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

Let's do the `partial fraction decomposition

#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))#

Therefore,

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

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

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

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

So,

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

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

#=-ln(∣x∣)+1/2ln(∣x+1∣)+ln(∣x-1∣)+C#