Answer
$${\theta ^2} + \theta + \sin \left( {2\theta + 1} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\left( {2\theta + 1 + 2\cos \left( {2\theta + 1} \right)} \right)} d\theta \cr
& = \int {2\theta } d\theta + \int {d\theta } + \int {2\cos \left( {2\theta + 1} \right)} d\theta \cr
& = {\theta ^2} + \theta + \int {2\cos \left( {2\theta + 1} \right)} d\theta \cr
& {\text{use the substitution method for the integral }}\int {2\cos \left( {2\theta + 1} \right)} d\theta {\text{. }} \cr
& u = 2\theta + 1,\,\,\,\,du = 2d\theta,\,\,\,\,\,d\theta = \frac{{du}}{2} \cr
& {\text{Then}}{\text{,}} \cr
& \int {2\cos \left( {2\theta + 1} \right)} d\theta = \int {2\cos u} \left( {\frac{{du}}{2}} \right) \cr
& = \int {\cos u} du \cr
& {\text{integrating }} \cr
& = \sin u + C \cr
& {\text{write in terms of }}\theta;{\text{ replace }}u = 2\theta + 1 \cr
& = \sin \left( {2\theta + 1} \right) + C \cr
& \cr
& {\text{Then we have}}{\text{,}} \cr
& = {\theta ^2} + \theta + \int {2\cos \left( {2\theta + 1} \right)} d\theta \cr
& = {\theta ^2} + \theta + \sin \left( {2\theta + 1} \right) + C \cr} $$
