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

Question

1224 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-08T19:19:43

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

We should solve the equation:

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

Putting everithing with the samw denominators:

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

Then we can eliminate denominators:

#x^2-1=(Ax^2+A)+(Bx^2+Cx)#

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

The we have three equations to solve:

Terms in #x^2: 1=A+B#

Terms in #x: 0=C#

Terms in #1: -1=A#

The solution is: A=-1, B=2, C=0.

So,

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