∫ arctg(5x) dx = {u=5x; du=5dx} = 1/5 ∫ arctg u * du =
v = u; dv = 1; du=1/(u^2+1)
= 1/5 (u*arctg u - ∫ u/(u^2+1) du) = 1/5(u*arctg u - 1/2 ∫ d(u^2)/(u^2+1))=
=(-2ln|u| + 1 + 2u* arctg u)/10 + C, где u=5x
∫tg x dx = ∫ sinx/cosx dx = -∫d(cosx)/cosx = -ln|cosx| + C