Evaluate

$\int \frac{(\sin x+\cos x)}{\sin ^{4} x+\cos ^{4} x} d x$

$=\int \frac{(\sin x+\cos x)}{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x} d x$

$=\int \frac{(\sin x+\cos x)}{1-2 \sin ^{2} x \cos ^{2} x} d x$

$=\int \frac{2(\sin x+\cos x)}{2-4 \sin ^{2} x \cos ^{2} x} d x$

$=\int \frac{2(\sin x+\cos x)}{2-\sin ^{2} 2 x} d x$

Let $\sin x-\cos x=t$,

$(\cos x+\sin x) d x=d t$

$=\int \frac{2}{2-\left(1-t^{2}\right)^{2}} d t$

$=\int \frac{2}{\left(\sqrt{2}-1+t^{2}\right)\left(\sqrt{2}+1-t^{2}\right)} d t$

$=\frac{1}{\sqrt{2}} \int\left(\frac{1}{\left(\sqrt{2}+1+t^{2}\right)}-\frac{1}{\left(\sqrt{2}-1-t^{2}\right)}\right) d t$

$=\frac{1}{\sqrt{2}} \int\left(\frac{1}{\left(\sqrt{2}+1+t^{2}\right)}\right) d t-\frac{1}{\sqrt{2}} \int\left(\frac{1}{\left(\sqrt{2}-1-t^{2}\right)}\right) d t$

$=\frac{1}{\sqrt{2}} \int\left(\frac{1}{\left(\left((\sqrt{\sqrt{2}+1)})^{2}+t^{2}\right)\right.}\right) d t-\frac{1}{\sqrt{2}} \int\left(\frac{1}{\left(((\sqrt{\sqrt{2}-1}))^{2}-t^{2}\right)}\right) d t$

$=\frac{1}{\sqrt{2}}\left[\frac{1}{2 \sqrt{\sqrt{2}+1}} \log \left|\frac{t-\sqrt{\sqrt{2}+1}}{t+\sqrt{\sqrt{2}+1}}\right|\right]-\frac{1}{\sqrt{2}}\left[\frac{1}{\sqrt{\sqrt{2}-1}} \tan ^{-1}\left(\frac{t}{\sqrt{\sqrt{2}-1}}\right)\right]+c$