Calculus: Early Transcendentals, Chapter 5, 5.4, Section 5.4, Problem 16

You need to find the indefinite integral, hence, you need to remember that sec t = 1/(cos t) , such that:
int (sec t)(sec t + tan t) dt = int (1/(cos t))(1 + sin t)/(cos t) dt
int (sec t)(sec t + tan t) dt = int 1/(cos^2 t) dt + int (sint)/(cos^2 t) dt
int (sec t)(sec t + tan t) dt = tan t + int (sint)/(cos^2 t) dt
You need to solve for the indefinite integral int (sint)/(cos^2 t) dt using the substitution cos t = u => -sint dt = du.
int (sint)/(cos^2 t) dt = int (-du)/(u^2) = 1/u + c
Replacing back cost for u yields:
int (sint)/(cos^2 t) dt = 1/(cos t) + c
Hence, evaluating the indefinite integral, yields int (sec t)(sec t + tan t) dt = tan t + 1/(cos t) + c = (sin t+1)/(cos t) + c