How to integrate this?

Question

(2 cos u - 2/u + 3 sin u) du?

Answer 1

ʃ[2 cosu - (2/u) + 3 sinu]du

Recall the formulae : ʃ(1/x) dx = log(x) + C

ʃ sinx dx = - cosx + C

ʃ cosx dx = sinx + C

ʃ[2 cosu - (2/u) + 3 sinu]du = ʃ2cosu du - ʃ(2/u)du + ʃ3 sinu du

= 2 ʃcosu du - 2ʃ(1/u)du + 3ʃ sinu du

= 2 sinu - 2 logu - 3 cosu + C.