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Topic:
Compute integral
Replies:
7
Last Post:
May 2, 2014 1:36 PM



Virgil
Posts:
8,833
Registered:
1/6/11


Re: Compute integral
Posted:
May 2, 2014 1:36 PM


In article <28a12e85cdc942ef994abd5dac1e9c72@googlegroups.com>, cootercrew@gmail.com wrote:
> On Friday, May 2, 2014 12:50:45 AM UTC4, coote...@gmail.com wrote: > > Compute the integral from 0 to 1 of ln(1+x)/(1+x^2)dx. Hint: Use the > > substitution y+1=2/(x+1) > > > > > > > > I'm pretty sure I need to get this in the form of the integral of ln u du. > > So I use the substitution given and let u = 2/(x+1) but when I compute the > > derivative of this I get [(x+1)*0  2(1)]/(1+x)^2 = 2 /(1+x)^2 which does > > not equal 1/(1+x^2) like I need. I didnt get too far trying integration by > > parts but maybe that is the way to go? I doubt it because of the hint so > > pretty sure I am just missing something getting it in the form of ln u du. > > > > > > > > Any help is appreciated! Thanks! This is just a practice problem for a > > final and not something I have to turn in. > > I haven't looked through all the comments closely yet but to clear things up > I did mean the integral of > ln(1+x) >  > (1+x^2) > > Also, this is a practice problem for a Real Analysis II class. I wish I as > still in Calc II. > Thanks for the replies!
Performing the indicated substitution, y+1=2/(x+1),
On Integral(0,1,ln(1+x)/(1+x^2),dx)
produces Integral(0,1, (ln(2)/(1+y^2),dy)  Integral(0,1,( ln(1+y)/ (1+y^2),dy)
The first of these two is easy and the second is equal to the original. 



