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,

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

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

$=-ln(∣x∣)+1/2ln(∣x+1∣)+ln(∣x-1∣)+C$

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