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

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

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

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