Answer
$$\frac{1}{{2\pi }}\left( {\pi x + \ln \left| {\cos \pi x + \sin \pi x} \right|} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{1 + \tan \pi x}}dx} \cr
& {\text{Let }}u = \pi x,{\text{ }}du = \pi dx,{\text{ }}dx = \frac{1}{\pi }du \cr
& {\text{Substituting}} \cr
& \int {\frac{1}{{1 + \tan \pi x}}dx} = \int {\frac{1}{{1 + \tan u}}\left( {\frac{1}{\pi }} \right)du} \cr
& = \frac{1}{\pi }\int {\frac{1}{{1 + \tan u}}du} \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {\frac{1}{{1 \pm \tan u}}du = \frac{1}{2}\left( {u \pm \ln \left| {\cos u \pm \sin u} \right|} \right) + C,{\text{ then}}} \cr
& = \frac{1}{{2\pi }}\left( {u + \ln \left| {\cos u + \sin u} \right|} \right) + C \cr
& {\text{Write in terms of }}u,{\text{ let }}u = \pi x \cr
& = \frac{1}{{2\pi }}\left( {\pi x + \ln \left| {\cos \pi x + \sin \pi x} \right|} \right) + C \cr} $$
