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

by carlosbgois
Tags: dx or sqrtd2, integral
 P: 49 Hi there. Evaluating the expression $\int\frac{dx}{\sqrt{x^{2}+y^{2}}}$ I can get to the result $ln(\frac{x+\sqrt{x^{2}+y^{2}}}{d})$, but in my book it goes from this directly to $ln (x+\sqrt{x^{2}+y^{2}})$, a result wolframalpha says is valid for 'restricted $x$ values'. What does it mean? What are those restricted values? Why? Many thanks.
P: 606
 Quote by carlosbgois Hi there. Evaluating the expression $\int\frac{dx}{\sqrt{x^{2}+y^{2}}}$ I can get to the result $ln(\frac{x+\sqrt{x^{2}+y^{2}}}{d})$, but in my book it goes from this directly to $ln (x+\sqrt{x^{2}+y^{2}})$, a result wolframalpha says is valid for 'restricted $x$ 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 $-\ln d$.

The question is: where did you get the constant $d$ from??

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

DonAntonio
 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)

 Related Discussions Calculus & Beyond Homework 5 General Math 13 General Math 6 Calculus & Beyond Homework 4 Calculus & Beyond Homework 5