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April 16, 2014

April 16, 2014

Posted by **sarah** on Sunday, January 27, 2008 at 2:49pm.

=-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.

- calc -
**Damon**, Sunday, January 27, 2008 at 3:49pmI 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) ]

- calc -
**drwls**, Sunday, January 27, 2008 at 3:51pmTo 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.

- calc -
**Damon**, Sunday, January 27, 2008 at 3:58pmSorry

ln (x+1)(x-1)

which is

ln(x^2-1)

- calc -
**Damon**, Sunday, January 27, 2008 at 3:59pmWhich I should have seen in the first place.

- calc -
**sarah**, Sunday, January 27, 2008 at 4:18pmtook ya long enuff

- calc -
**Damon**, Sunday, January 27, 2008 at 4:26pmYeah, but I am really old :)

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