ulisis-
The integral of (1/baomer) dbaomer would be ln |baomer|, (don’t forget those absolute value signs, unless you specify that baomer is strictly positive).
However the integral of 1/(ln(10)*baomer) dbaomer would in fact be Log(BaOmer) since Log (base 10) BaOmer = Ln (Baomer) / LN (10), and stam Logs (no specified base) are base 10.
=> Log (base 10) BaOmer = Log(BaOmer)