I bet you are! (plus or minus a twist)oh I see... so [tex]d(r^2)[/tex] just basically means taking the derivative of [tex]r^2[/tex] with respect with r?
I am not familiar with that notation
thanks for clearing that upI bet you are! (plus or minus a twist)
Consider [tex]I = \int (u + a^2)^{-\frac{3}{2}} \, du.[/tex]
Let [tex]u = r^2,[/tex] so [tex]du = 2r\, dr.[/tex] Then
[tex]I = \int (r^2 + a^2)^{-\frac{3}{2}} \, 2r \, dr\, \text{!}[/tex]