yes, you are right...almost...
u= ln (39x)
but ln (ab)=lna + lnb
u= ln39 + lnx
du= 0+1/x dx
Silly question here, but I am not getting a certain part in the solution in a problem.
Evaluate the integral of [(sin(ln(39x)))/x]dx. We are supposed to take u=ln(39x), and get du=(1/x)dx, and the then whole thing is simply solved using integration formulas. But my confusion is in the du=(1/x)dx. If u=ln(39x)m shouldnt du=39(1/39x)dx ?? I did this and got an extra -1/39 in my answer, but the solution doesnt have that as they just took (1/x)dx. I wonder why? Am I missing something here?
2 answers
Ohhh yeah! Got it! Thanks!
2 answers