How to evaluate $\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx$? How can one find $$\int  \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx?$$

Let
\begin{align}
 I &= \int {\frac{{\sin x + \cos x}}{{\sin ^4 x + \cos ^4 x}}dx}  \\ 
  &= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx}  + \int {\frac{{\cos x}}{{\sin ^4 x + \cos ^4 x}}dx}  \\ 
  &= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx}  + \int {\frac{{\sin \left( {{\textstyle{\pi  \over 2}} - x} \right)}}{{\sin ^4 x + \sin ^4 \left( {{\textstyle{\pi  \over 2}} - x} \right)}}dx} ,\,\,\,\,\,\, (\text{since}\,\, 
{\sin \left( {{\textstyle{\pi  \over 2}} - x} \right)}=\cos x)
\end{align}
Substituting $u=\frac{\pi}{2}-x$ in the second integral we get
\begin{align}
I&= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx}  - \int {\frac{{\sin u}}{{\sin ^4 \left( {{\textstyle{\pi  \over 2}} + u} \right) + \sin ^4 u}}du} 
\\
 &= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx}  - \int {\frac{{\sin u}}{{\cos ^4 u + \sin ^4 u}}du} ,\,\,\,\,\,\,\,\,\,\,\,\, (\text{since}\,\, 
{\sin \left( {{\textstyle{\pi  \over 2}} + u} \right)}=\cos u)
\\
&=0
\end{align}
