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

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Virgil

Posts: 8,833
Registered: 1/6/11
Re: Compute integral
Posted: May 2, 2014 1:36 PM
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In article <28a12e85-cdc9-42ef-994a-bd5dac1e9c72@googlegroups.com>,
cootercrew@gmail.com wrote:

> On Friday, May 2, 2014 12:50:45 AM UTC-4, 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.
--





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