Question

is the integral of 2x/(x^3-x)
=-1/2ln(x+1)+1/2ln(x-1)

I got this by doing this:
(\=integral sign)
2\1/(x^2-1)
2\1/(x+1)(x-1)
=A/(x+1)+B/(x-1)
=Ax-A+Bx+B=1
Ax+Bx=0
B-A=1
B=1+A
(1+A)x+Ax=0
x+Ax+Ax=0
1+AB=0
2A=-1
A=-1/2
B+1/2=1, B=1/2
etc.

Answer 1

I differ by a factor of 2

2[ A/(x+1) +B/(x-1) ]
A = -1/2, B = +1/2
so
2 int [ -.5 dx/(x+1) + .5 dx /(x-1) ]

1 int [ dx/(x+1) + dx/(x-1) ]

ln (x+1) + ln (x-1)
which is
ln[ (x+1)/(x-1) ]

Answer 2

To see if your answer is correct, take the derivative. Itis
-(1/2)/(x+1) +(1/2)/(x-1)
= [-(1/2)(x-1) + (1/2)(x+1)]/(x^2 -1)
= 1/(x^2-1)
Your riginal integrand is equal to 2/(x^2-1), so your answer seems to be off by a factor of 2.

Answer 3

Sorry
ln (x+1)(x-1)
which is
ln(x^2-1)

Answer 4

Which I should have seen in the first place.

Answer 5

took ya long enuff

Answer 6

Yeah, but I am really old :)