Integral of (cosx-cos^2x+1)/sinxcosx dx

i know that the answer is ln(cscx-cotx) + ln(secx) + C
but i have no idea how to get there. thanks!

Note that you can write:
cos(x) - cos^2(x) + 1 = cos(x) + [1 - cos^2(x)]
= cos(x) + sin^2(x), since sin^2(x) + cos^2(x) = 1.

So, we have:
∫ [cos(x) - cos^2(x) + 1]/[sin(x)cos(x)] dx = ∫ [cos(x) + sin^2(x)]/[sin(x)cos(x)] dx
= ∫ {cos(x)/[sin(x)cos(x)] + sin^2(x)/[sin(x)cos(x)]} dx
= ∫ [1/sin(x) + sin(x)/cos(x)] dx
= ∫ [csc(x) + tan(x)] dx
= ∫ csc(x) dx + ∫ tan(x) dx
= ln|csc(x) - cot(x)| + ln|sec(x)| + C, from the standard integrals.

I hope this helps!

-Log[Cos[x/2]] - Log[Cos[x]] + Log[Sin[x/2]]