∫[(√tanx)+1]/cos²x dx
=∫sec²x·[(√tanx)+1] dx
=∫[(√tanx)+1] d(tanx)
=2/3·(tanx)^(3/2)+tanx+C=2/3·(tanx)^(3/2)+tanx+C????∫[(√tanx)+1] d(tanx)=∫[(tanx^^(1/2)+1] d(tanx)=2/3·(tanx)^(3/2)+tanx+C