`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` .

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