Finding the integral of 1/x(x+1)(x-1)(x²+1)

[I like Serena]

nevermind, i wrote it down opposite from you :)

okay so now i have
1/4 * [ln(u)-ln(u+1)] between 1/3 and 1/2, right?

Yep!

EDIT: err, no. The limits are not right.

 

[Shannabel]

Yep!

EDIT: err, no. The limits are not right.

what's wrong with the limits?
i went on and got the right answer:

1/4*[1/u-1/(u+1)]between 1/3 and 1/2
=1/4(lnu-ln(u+1))
=1/4(lnu/(u+1))
when x=1/2, u=-15/16
when x=1/3, u=-80/81
= 1/4 [ln((-15/16)/(1/16))-ln((-80/81)/(1/81))]
= 1/4 [ln(-15)-ln(-80)]
=1/4 [ln(15/80)]
= 1/4ln(3/16)

:)

 

[Bohrok]

what's wrong with the limits?
i went on and got the right answer:

1/4*[1/u-1/(u+1)]between 1/3 and 1/2
=1/4(lnu-ln(u+1))
=1/4(lnu/(u+1))
when x=1/2, u=-15/16
when x=1/3, u=-80/81

The limits 1/3 and 1/2 would've been wrong if you used them while the integral was in terms of u, but you changed them right here and those are the correct limits; you didn't tell us you changed them in the previous post

 

[I like Serena]

What Bohrok said!