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Integration- by sub??

Find integral of
[(1-x)/(1+x)] dx
(edited 8 years ago)
Reply 1
Original post by Ano123
Find integral of
[(1-x)/(1+x)] dx


my first attempt would be 2/(1+x) = cosh2u
Reply 2
Original post by Ano123
Find integral of
[(1-x)/(1+x)] dx

Try x=cosux=\cos u substitution and then use trig identities.
Reply 3
How about u = 1-x ?
That will turn your integrand into u/u-2 which I assume can be simplified by long division?
Reply 4
Original post by oShahpo
How about u = 1-x ?
That will turn your integrand into u/u-2 which I assume can be simplified by long division?

But you still have the square root..
Reply 5
Original post by notnek
But you still have the square root..


That is the tiniest square root I have never seen in my life..
Reply 6
Original post by oShahpo
That is the tiniest square root I have never seen in my life..

They're the deadliest kind.
Reply 7
Try x=cos2θ
Simplifies very nicely then.
Reply 8
Even better, try multiplying up and down by root(1-x), you'll get two fractions which you can then integrate by substitution.
Reply 9
Original post by oShahpo
Even better, try multiplying up and down by root(1-x), you'll get two fractions which you can then integrate by substitution.


Can you show me what that leads to, I'm not sure that will work.
Reply 10
Original post by Ano123
Find integral of
[(1-x)/(1+x)] dx


Original post by TeeEm
my first attempt would be 2/(1+x) = cosh2u


actually the quickest way to do this is to multiply by (1-x), top and bottom under the root
simplify
split the fraction
and you get 2 very easy antiderivatives
(edited 8 years ago)
Original post by Ano123
Can you show me what that leads to, I'm not sure that will work.


It does, you get two fractions. One can be done by a very simple substitution and the other can be done by knowledge of integration of Inverse Hyperbolics.
Reply 12
Original post by oShahpo
It does, you get two fractions. One can be done by a very simple substitution and the other can be done by knowledge of integration of Inverse Hyperbolics.

Indeed after trying it, it does appear to be the easiest way, although using a trig sub can work as well. :biggrin:

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