- #1

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**infinity**

[tex]\int\sqrt{x}/(x^2 + 1)dx[/tex]

Thank you very much

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- Thread starter Bachelier
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- #1

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[tex]\int\sqrt{x}/(x^2 + 1)dx[/tex]

Thank you very much

- #2

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Try Integration By Parts

[tex]\int[/tex] u dv = uv - [tex]\int[/tex] v du

[tex]\int[/tex] u dv = uv - [tex]\int[/tex] v du

- #3

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Unfortunately I get stuck in the interval [0,1].

- #4

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substitute u = sqr(x), then use partial fractions

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Dick

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Dick

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Unfortunately I get stuck in the interval [0,1].

You can't really 'integrate' using the comparison method, but you can show the integral exists and find a bound on it. If that's all you want to do, it's a lot easier than actually doing the integration. To deal with the interval [0,1] you had the right idea when you looked at the graph and observed it's approximately between 0 and 1/2. You could just compute the upper bound by finding the derivative, looking for critical points etc. If you do you'll find it's actually a bit larger than 1/2.

- #8

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[tex]\frac{\pi}{\sqrt{2}}[/tex]

- #9

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[tex]\frac{\pi}{\sqrt{2}}[/tex]

I am not yet familiar with contour integration. But thank you for the heads-up. I am going to run a google search for it.

- #10

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You can't really 'integrate' using the comparison method, but you can show the integral exists and find a bound on it. If that's all you want to do, it's a lot easier than actually doing the integration. To deal with the interval [0,1] you had the right idea when you looked at the graph and observed it's approximately between 0 and 1/2. You could just compute the upper bound by finding the derivative, looking for critical points etc. If you do you'll find it's actually a bit larger than 1/2.

Thank you Dick...Your answer helped a lot in directing me towards the correct answer.

in [0,1], I get a maxima of [tex]1/\sqrt{3}[/tex]. The graph interestingly goes beyond y=1/2 but never above y= 0.6, making the area less than or equal to [tex]1/\sqrt{3}[/tex] in the said interval.

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Dick

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Indeed..Thanks for the editing. :)

You are absolutely correct, but if we wanted to narrow down the area, we can use the graph to calculate the maximum value of y, here y= f(1/sqrt(3))=0.59 and since x=1, then the area under the graph is smaller or equal to x*y = 0.59...

What do you think????

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Dick

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Dick, thank you for coaching me through this problem. It allowed me to review and practice some concepts. I hope we can repeat this in the future with more problems.

On another note, the greatest discovery for me was the engine called Wolfram Alpha mentioned by Waht. This discovery by itself was worth the thread. Although Waht decided to delete the link, I am still thankful. I think this is a great tool for someone who wants to learn and understand mathematics, and not just do homework or get an A or B in a class.

I have Maple and Mathematica 7 installed on my computer, yet I found alpha a lot smoother and more fun.

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Dick

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