\begin{array}{rl}

\displaystyle\therefore \int_0^1 \frac{(1-y)^3}{(1+y)^5} \arctan y \, dy & = \displaystyle\int_1^0 \frac{x^3}{(1+y)^2} \arctan \left( \frac{1-x}{1+x} \right) \times -\frac{1}{2}(1+y)^2 \, dx \\ \br \\

& \displaystyle = \int_1^0 - \frac12 x^3 \arctan \left( \frac{1-x}{1+x} \right) \, dx \\ \br \\

& \displaystyle = \frac12 \int_0^1 x^3 \arctan \left(\frac{1-x}{1+x} \right) \, dx \\ \br \\

& \displaystyle = \frac12 \left( \frac{\pi}{16} - \frac{1}{6} \right) \\ \br \\

& \displaystyle = \boxed{\frac{\pi}{32} - \frac{1}{12}} \end{array}