Answer:
1/2 ㏑(sec (2x + 9)) + c ........ absolute sec (2x +9)
Explanation:
∫tan (2x + 9) dx
let u = 2x + 9
du/dx = 2
dx = du/2 = (1/2) du
now we have ∫tan u (1/2 du)
= 1/2 ∫tan u du
= 1/2 ㏑ (sec u) ...... absolute
=1/2 ㏑(sec (2x + 9)) + c ........ absolute