Calculate Integral [(1+(tgx)^2)/tg x]dx and specify the method.

First of all, we'll notice that 1+(tgx)^2 = 1/(cos x)^2,
from the fundamental formula of trigonometry:

(sin x)^2 +
(cos x)^2 = 1

(sin x)^2/(cos x)^2 + 1 = 1/(cos
x)^2

(tg x)^2 + 1 = 1/(cos
x)^2

Int [(1+(tgx)^2)/tg x]dx=Int dx/(tg x)(cos
x)^2

Now, we can choose the method of
substitution.

tg x = t, so, differentiating, we'll
have:

 dx/(cos x)^2 =
dt

Int (1/t)dt = ln t + C = ln (tg x) +
C