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# integral of 1/(A^2+B^2)^3/2 (integrate over B)

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Dear mathlovers,

How do you evaluate this integral:$$\int{1}/{((A^2+B^2)^\frac{3}{2}) }dB$$
You have to use substitution somehow, but I don't know how to arrive at the point that you can.
Could you help me?

Kind regards,
David

Guest Oct 28, 2015

#1
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As below:

.

Alan  Oct 28, 2015
Sort:

#1
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As below:

.

Alan  Oct 28, 2015
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That was fantastic! Thank you! :)

Guest Oct 28, 2015
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You're welcome!

Alan  Oct 28, 2015
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Very nice, Alan....!!!!

CPhill  Oct 28, 2015
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Alan: I plugged it into "WolframAlpha" engine!!. This is the result it gives. I'm not very good at Integral Calculus, so I cannot judge if the two are the same or not. They certainly look different:

http://www.wolframalpha.com/input/?i=integral+of+1%2F%28A%5E2%2BB%5E2%29%5E3%2F2+%28integrate+over+B%29.

If you cannot see the above link, here is their answer:1/(2 (A^2+B^2)^3)

Guest Oct 28, 2015
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WolframAlpha hasn't interpreted your syntax the way you intended. See

http://www.wolframalpha.com/input/?i=integrate+1%2F%28A%5E2%2BB%5E2%29%5E%283%2F2%29+with+respect+to+B

Also notice that I've put brackets around the 3/2 power.

Alan  Oct 28, 2015
edited by Alan  Oct 28, 2015
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Hi David,

If you are going to ask a few good questions like this why don't you join up so we can get to know you.

It is easier than posting as a guest all the time too.  You only need to choose a username and a password, that is it!

Thanks Alan for that fabulous answer.

Chris sent the problem in a private message to make sure  that i saw it, and that i would include it in my next wrap.  :)