Posted by sarah on Sunday, January 27, 2008 at 2:49pm.
is the integral of 2x/(x^3x)
=1/2ln(x+1)+1/2ln(x1)
I got this by doing this:
(\=integral sign)
2\1/(x^21)
2\1/(x+1)(x1)
=A/(x+1)+B/(x1)
=AxA+Bx+B=1
Ax+Bx=0
BA=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 3:49pm
I differ by a factor of 2
2[ A/(x+1) +B/(x1) ]
A = 1/2, B = +1/2
so
2 int [ .5 dx/(x+1) + .5 dx /(x1) ]
1 int [ dx/(x+1) + dx/(x1) ]
ln (x+1) + ln (x1)
which is
ln[ (x+1)/(x1) ]

calc  drwls, Sunday, January 27, 2008 at 3:51pm
To see if your answer is correct, take the derivative. Itis
(1/2)/(x+1) +(1/2)/(x1)
= [(1/2)(x1) + (1/2)(x+1)]/(x^2 1)
= 1/(x^21)
Your riginal integrand is equal to 2/(x^21), so your answer seems to be off by a factor of 2.

calc  Damon, Sunday, January 27, 2008 at 3:58pm
Sorry
ln (x+1)(x1)
which is
ln(x^21)

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

calc  sarah, Sunday, January 27, 2008 at 4:18pm
took ya long enuff

calc  Damon, Sunday, January 27, 2008 at 4:26pm
Yeah, but I am really old :)
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