Answer
$$\int \frac{1+sinx}{1+cosx}dx=-cotx+ln\left |cscx -cotx\right |+cscx-ln\left | sinx \right |+C$$
Work Step by Step
$$\int \frac{1+sinx}{1+cosx}dx=\int \frac{1+sinx}{1+cosx}\frac{1-cosx}{1-cosx}dx$$
$$=\int \frac{1+sinx-cosx-sinx\,cosx}{sin^{2}x}dx$$
$$=\int (csc^{2}x+csc\,x-cscx\,cotx-cotx)dx$$
$$=-cotx+ln\left |cscx -cotx\right |+cscx-ln\left | sinx \right |+C$$
