Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 7

int (x^2+2x)cosx dx
To evaluate, apply integration by parts int udv = uv - int vdu .
So let
u = x^2+2x
and
dv = cosx dx
Then, differentiate u and integrate dv.
du = (2x + 2)dx
and
v = int cosx dx = sinx
Plug-in them to the formula of integration by parts. So the integral becomes:
int (x^2+2x)cosx dx
= (x^2+2x)sinx - int sinx * (2x + 2)dx
= (x^2 + 2x)sinx - int (2x + 2)sinx dx
To take the integral of (2x + 2sinx)dx, apply integration by parts again.
So let
u_2 = 2x + 2
and
dv_2 = sinx dx
Differentiate u_2 and integrate dv_2.
du_2 = 2dx
and
v_2 = -cosx
So the integral becomes:
= (x^2+2x)sinx - [ (2x + 2)*(-cosx) - int -cosx * 2dx]
=(x^2+2x)sinx - [-(2x + 2)cosx + 2int cosx dx]
=(x^2+2x)sinx - [-(2x + 2)cosx + 2sinx]

= (x^2+2x)sinx +(2x +2)cosx -2sinx
= (x^2+2x - 2)sinx + (2x + 2)cosx
Since the given is indefinite integral, add C.
= (x^2+2x - 2)sinx + (2x + 2)cosx + C

Therefore, int (x^2+2x)cosx dx = (x^2+2x-2)sinx= (x^2+2x - 2)sinx + (2x + 2)cosx + C .