How do you find integral of ##((secxtanx)/(secx-1))dx##?

Another way of doing this is to consider that ##d(secx) = secxtanx*dx##

That is, the derivative of ##secx## is ##secxtanx##

##=> int(secxtanx)/(secx - 1)dx = int d(secx)/(secx - 1)##

This is the same as letting ##u = secx##

We then have,

##int(du)/(u - 1) = ln(u - 1) + C = ln(secx - 1) + C##