Question

Partial fraction...evaluate the integral of (x^2+2x-1)/(x^3-x)

Answer 1

We can factor the denominator as x(x^2 - 1). Since the denominator can be expressed as there distinct linear factors, we now that there exist numerators A, B, and C such that\[\frac{ x^2 + 2x - 1 }{ x(x^2 - 1) } = \frac{ A }{ x } + \frac{ B }{ x - 1 }+\frac{ C }{ x + 1}\]

Answer 2

Multiply everything by the denominator on the left hand side and observe that\[x^2 + 2x - 1 = A(x^2 -1) + B(x^2 + x) + C(x^2 - x)\]

Answer 3

Thus, \[(A + B + C) x^2 + (B -C)x - A = x^2 + 2x - 1\]

Answer 4

From this we can see that A is 1. Plugging in and solve the linear system we find that B = 1 and C = -1

Answer 5

Does this make sense so far?

Answer 6

Yeah, I get it up until now

Answer 7

So what are we left with? Plug A,B and C in our first equation. Then,\[\frac{ x^2 + 2x - 1 }{ x^3 - x} = \frac{ 1 }{ x } + \frac{ 1 }{ x -1 } - \frac{ 1 }{ x + 1 }\]

Answer 8

Now we can integrate quite easily.

Answer 9

So it's lnx+ln(x-1)-ln(x+1) after it's been integrated

Answer 10

Yup