What Are the Restricted x Values for the Integral of dx/(sqrt(d^2 + x^2))?

  • Level: Undergrad 
  • Thread starter Thread starter carlosbgois
  • Start date Start date
  • Tags Tags
    Integral
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 3K views
carlosbgois
Messages
66
Reaction score
0
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.
 
Physics news on Phys.org
carlosbgois said:
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
 
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)