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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

 Oct 28, 2015

Best Answer 

 #1
avatar+28182 
+15

As below:

 

integration

.

 Oct 28, 2015
 #1
avatar+28182 
+15
Best Answer

As below:

 

integration

.

Alan Oct 28, 2015
 #2
avatar
+5

That was fantastic! Thank you! :)

 Oct 28, 2015
 #3
avatar+28182 
+5

You're welcome!

 Oct 28, 2015
 #4
avatar+104793 
+5

Very nice, Alan....!!!!

 

 

cool cool cool

 Oct 28, 2015
 #5
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+5

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)

 Oct 28, 2015
 #6
avatar+28182 
+5

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.

 Oct 28, 2015
edited by Alan  Oct 28, 2015
 #7
avatar+105606 
0

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.  :)  

Chris worked through your answer  and was very impressed by your logic.    laugh

Thanks for sending it to me Chris.   :)

 Oct 29, 2015

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