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Integral of dx/(sqrt(d^2+x^2))

by carlosbgois
Tags: dx or sqrtd2, integral
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carlosbgois
#1
Mar30-12, 11:43 AM
P: 49
Hi there. Evaluating the expression [itex]\int\frac{dx}{\sqrt{x^{2}+y^{2}}}[/itex] I can get to the result [itex]ln(\frac{x+\sqrt{x^{2}+y^{2}}}{d})[/itex], but in my book it goes from this directly to [itex]ln (x+\sqrt{x^{2}+y^{2}})[/itex], a result wolframalpha says is valid for 'restricted [itex]x[/itex] values'. What does it mean? What are those restricted values? Why?

Many thanks.
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DonAntonio
#2
Mar30-12, 12:50 PM
P: 606
Quote Quote by carlosbgois View Post
Hi there. Evaluating the expression [itex]\int\frac{dx}{\sqrt{x^{2}+y^{2}}}[/itex] I can get to the result [itex]ln(\frac{x+\sqrt{x^{2}+y^{2}}}{d})[/itex], but in my book it goes from this directly to [itex]ln (x+\sqrt{x^{2}+y^{2}})[/itex], a result wolframalpha says is valid for 'restricted [itex]x[/itex] values'. What does it mean? What are those restricted values? Why?

Many thanks.


Well, since this is indefinite integration both the results are correct as their difference is just the constant [itex]-\ln d[/itex].

The question is: where did you get the constant [itex]d[/itex] from??

The result is valid for any values of [itex]x,y, s.t. x^2+y^2\neq 0[/itex]

DonAntonio
carlosbgois
#3
Mar30-12, 12:54 PM
P: 49
Thank you. 'd' is actually a constant length (the distance from a point p to a disk, in an axis that goes through the center of the disk)


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