Thomas' Calculus 13th Edition

$${\theta ^2} + \theta + \sin \left( {2\theta + 1} \right) + C$$
\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}