How do you express $x/((x^2-1)(x-1))$ in partial fractions?

Question

14610 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-16T08:07:03

The answer is $=(-1/4)/(x+1)+(1/2)/(x-1)^2+(1/4)/(x-1)$

Let's factorise the denominator

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

Therefore,

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

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

Let's perform the decomposition into partial fractions

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

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

The denominators are the same, we compare the numerators

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

Let $x=-1$ , $=>$ , $-1=4A$ , $=>$ , $A=-1/4$

Let $x=1$ , $=>$ , $1=2B$ , $=>$ , $B=1/2$

Coefficients of $x^2$

$0=A+C$ , $=>$ , $C=-A=1/4$

Therefore,

$x/((x^2-1)(x-1))=(-1/4)/(x+1)+(1/2)/(x-1)^2+(1/4)/(x-1)$

How do you express x/((x^2-1)(x-1))x/((x^2-1)(x-1)) in partial fractions?