Answer
$$\frac{3}{8}\pi $$
Work Step by Step
$$\eqalign{
& \int_0^3 {\int_{y/3}^1 {\frac{1}{{1 + {x^4}}}} dxdy} \cr
& {\text{Switching the order of integration, the new region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant y \leqslant 3x,{\text{ }}0 \leqslant x \leqslant 1} \right\} \cr
& {\text{Therefore,}} \cr
& \int_0^3 {\int_{y/3}^1 {\frac{1}{{1 + {x^4}}}} dxdy} = \int_0^1 {\int_0^{3x} {\frac{1}{{1 + {x^4}}}} dydx} \cr
& = \int_0^1 {\left[ {\int_0^{3x} {\frac{1}{{1 + {x^4}}}} dy} \right]} dx \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^1 {\left[ {\frac{y}{{1 + {x^4}}}} \right]_0^{3x}} dx \cr
& = \int_0^1 {\left[ {\frac{{3x}}{{1 + {x^4}}} - \frac{0}{{1 + {x^4}}}} \right]} dx \cr
& = 3\int_0^1 {\frac{x}{{1 + {x^4}}}} dx \cr
& = \frac{3}{2}\int_0^1 {\frac{{2x}}{{1 + {{\left( {{x^2}} \right)}^2}}}} dx \cr
& {\text{Integrate}} \cr
& = \frac{3}{2}\left[ {{{\tan }^{ - 1}}\left( {{x^2}} \right)} \right]_0^1 \cr
& = \frac{3}{2}\left[ {{{\tan }^{ - 1}}\left( {{1^2}} \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right] \cr
& = \frac{3}{2}\left( {\frac{\pi }{4} - 0} \right) \cr
& = \frac{3}{8}\pi \cr} $$
