Answer
$$\frac{1}{8}\left( {4x - \ln \left| {\sin 4x + \cos 4x} \right|} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{1 + \cot 4x}}} dx \cr
& {\text{Let }}u = 4x,{\text{ }}du = 4dx,{\text{ }}dx = \frac{{du}}{4} \cr
& {\text{Substitute}} \cr
& \int {\frac{1}{{1 + \cot 4x}}} dx = \int {\frac{1}{{1 + \cot u}}} \left( {\frac{{du}}{4}} \right) \cr
& = \frac{1}{4}\int {\frac{1}{{1 + \cot u}}} du \cr
& {\text{Use a table of integrals with forms involving trigonometric}} \cr
& {\text{functions, from the table of integrals in the back of the book}} \cr
& {\text{Formula 72}} \cr
& \int {\frac{1}{{1 + \cot u}}} du = \frac{1}{2}\left( {u - \ln \left| {\sin u + \cos u} \right|} \right) + C \cr
& \frac{1}{4}\int {\frac{1}{{1 + \cot u}}} du = \frac{1}{8}\left( {u - \ln \left| {\sin u + \cos u} \right|} \right) + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}4x{\text{ for }}u \cr
& = \frac{1}{8}\left( {4x - \ln \left| {\sin 4x + \cos 4x} \right|} \right) + C \cr} $$
