Integrate cos(3x+1) dx

is it (sin(3x+1))/3 + c?? i'm confused ><

You are right. When in doubt, just perform the integration using a simple method. One such is shown next.

Put 3x+1 = t. So this means 3dx = dt and hance dx = dt/3

I = integral [cos(3x+1)dx]

After substitution, it becomes,
I = integral [cos(t) *(dt/3)]
I = (1/3) integral [cos(t) dt]
I = (1/3) *sin(t) + C, C = arbitrary constant
I = (1/3)*sin(3x+1) + C, substituting back for t

As revealed in your question im in the presents of a function within another function(composite function) thus it calls for the u-subtitution

let u = 3x+1
therefore du = 3dx
==> dx = du/3
Inte cosu du/3
1/3 inte cos u du

answer : I = 1/3 sin(3x+2)+ C
Your answer is precise :]

{sin(3x + 1)}/3

i guess it is. what is the confusion?

yup you are right be confident.