Answer
$$\frac{1}{2}\ln \left| {\sin \left( {{x^2} + 2x} \right)} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {x + 1} \right)\cot \left( {{x^2} + 2x} \right)} dx \cr
& {\text{substitute }}u = {x^2} + 2x \cr
& du = \left( {2x + 2} \right)dx \cr
& \frac{1}{2}du = \left( {x + 1} \right)dx \cr
& = \int {\cot u\left( {\frac{1}{2}} \right)} du \cr
& = \frac{1}{2}\int {\cot u} du \cr
& = \frac{1}{2}\int {\frac{{\cos u}}{{\sin u}}} du \cr
& {\text{find the antiderivative }} \cr
& = \frac{1}{2}\ln \left| {\sin u} \right| + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = {x^2} + 2x \cr
& = \frac{1}{2}\ln \left| {\sin \left( {{x^2} + 2x} \right)} \right| + C \cr} $$
