- #1

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fnInt( (Ln(x+1)/(x^2 +1)),x,0,1)

If you dont know that notation. it means Integral from 0 to 1 of Ln(x+1)/(x^2 +1)

OBVIOUSLY NO CALCULATORS... im looking for a solution, not an answer

Thanks

~rosie

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- Thread starter SNOOTCHIEBOOCHEE
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- #1

- 145

- 0

fnInt( (Ln(x+1)/(x^2 +1)),x,0,1)

If you dont know that notation. it means Integral from 0 to 1 of Ln(x+1)/(x^2 +1)

OBVIOUSLY NO CALCULATORS... im looking for a solution, not an answer

Thanks

~rosie

- #2

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I believe that if you try an integration by parts, with u=ln(x+1) and dv=1+x^2, that should get you stated--i believe will will have to do one more integration by parts and then some long division but i think that will get you to the end

- #3

HallsofIvy

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[tex]\frac{ln(x+1)}{x^2+1}[/tex]

which is what nate808 is asumming or

[tex]ln(\frac{x+1}{x^2+1})[/tex]

which is how I would read it?

- #4

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dv = 1 dx, and [tex]u = ln(\frac{x+1}{x^2+1})[/tex], and be prepared for alot of messy simplification and some partial fractions . Note the integral of [tex]1 /(x^2 + 1)[/tex] is [tex]tan^{-1} (x)[/tex] if it's the other integral then good luck to you.

- #5

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HallsofIvy said:

[tex]\frac{ln(x+1)}{x^2+1}[/tex]

which is what nate808 is asumming or

[tex]ln(\frac{x+1}{x^2+1})[/tex]

which is how I would read it?

I would guess, based on the parentheses that he used, that it should be:

[tex]\frac{ln(x+1)}{x^2+1}[/tex], in which case parts makes the most sense. Though, I would most likely take [tex]u=ln(x+1)[/tex] and [tex]dv=\frac{dx}{x^2+1}[/tex]. Integrating dv will most likely require either partial fractions or trig. substitution, though I can't honestly say that I've worked it out.

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- #6

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[tex] \int_{0}^{1}\frac{\arctan x}{1+x}dx=\allowbreak \mbox{\func{Catalan}}-\frac{1}{2}i \ \mbox{\func{dilog}}\left( \frac{1}{2}+\frac{1}{2}i\right) +\frac{1}{2}i \ \mbox{\func{dilog}}\left( \frac{1}{2}-\frac{1}{2}i\right) [/tex]

Daniel.

- #7

Hurkyl

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My next thought is complex analysis. Maybe there's some nice contour over which you can integrate that simplifies this. My first thought is to integrate it over a rectangle, two of whose edges are [0, 1] on the real axis, and [0, H] on the complex axis, then take the limit as H goes to infinity. I haven't tried working through the algebra though, I don't know if it will be helpful.

Note that if you try this approach, you either have to use a small semicircle to avoid the pole at

My third thought is to try a power series: you can find the power series for the integrand, and you can integrate that power series -- maybe you can recognize the resulting infinite sum.

Maybe I'll think about this a bit at work and have a better idea for you later.

- #8

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Widder shows how this integral is equal to -(pi^2)/8.

1) Substitute 1 - x^2 + x^4 - x^6... for the 1/(1-x^2) term

2) Interchange summation and integration

3) integrate (x^2k)*ln(x+1) by parts

4) Sum the infinite series

- #9

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lots of ingenious integration devices in there actually...

- #10

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factor x^2 + 1 = (x+i)(x-i) and use partial fraction A/x+i B/x-i

- #11

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lots of ingenious integration devices in there actually...

The book isn't free unfortunately.

- #12

Curious3141

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This is a problem from the 2005 Putnam competition (problem A5, to be precise). It can be solved with a choice of nifty substitutions :

Problems : http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2005.pdf [Broken]

Solutions : http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2005s.pdf [Broken]

Problems : http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2005.pdf [Broken]

Solutions : http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2005s.pdf [Broken]

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- #13

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Integral from 0 to 1 of ln(x+1)/(x^2 +1) dx isn't that hard actually but I only know one way to do it.

1/(x^2+1) = (arctg x)' ... (ie the derivate of arctan x) tg(x)=tan(x)

you change the variable and the integral becomes

integral from 0 to pi/4 of ln(1+tg(x)) dx

by making the following substitution you should get it done:

x=pi/4-u

dx=-du

the integral becomes

- integral from pi/4 to 0 of ln(1+tg(pi/4-u)) du =

integral from 0 to pi/4 of ln(1+((1-tg(u)/(1+tg(u))) du =

integral from 0 to pi/4 of ln(2/(1+tg(u)) du =

integral from 0 to pi/4 of ln2 du - integral from 0 to pi/4 of ln(1+tgu) du

and you're back from where you started.

That means our integral I is

I=pi/4*ln2-I => 2I=pi/4*ln2 => I=pi/8*ln2

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