${\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\sin x+\cos x}{{\cos }^{2}x+{\sin }^{4}x}dx=}$

$I={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\sin xdx}{{\cos }^{2}x+{\left(1-{\cos }^{2}x\right)}^2}}+{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\cos xdx}{1-{\sin }^{2}x+{\sin }^{4}x}}$ 

$t=\cos x$  in the first integral and $t=\sin x$  in the second integral gives

$I={\displaystyle \underset{\frac{1}{\sqrt{2}}}{\overset{1}{\int }}\frac{dt}{1-t^2+t^4}+{\displaystyle \underset{0}{\overset{\frac{1}{\sqrt{2}}}{\int }}\frac{dt}{1-t^2+t^4}}}$

   $={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1-t^2+t^4}}$ $={\displaystyle \underset{0}{\overset{1}{\int }}\frac{t^{-2}}{t^2+\frac{1}{t^2}-1}}$

   $=\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{2\left(\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}-1}=\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{1+\frac{1}{t^2}-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}-1}dt}}$

  $=\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{1+\frac{1}{t^2}}{{\left(t-\frac{1}{t}\right)}^2+1}dt-\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{1-\frac{1}{t^2}}{{\left(t+\frac{1}{t}\right)}^2-3}dt}}$

  $=\frac{1}{2}{\left[\tan { }^{-1}\left(t-\frac{1}{t}\right)\right]}_0^1-\frac{1}{4\sqrt{3}}\ln {\left[\frac{t+\frac{1}{t}-\sqrt{3}}{t+\frac{1}{t}+\sqrt{3}}\right]}_0^1$

$=\frac{\pi }{4}+\frac{1}{2\sqrt{3}}\ln \left(2+\sqrt{3}\right)$