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Let $I = \int_{0}^{\pi/2}\frac{x}{\sin x+\cos x}dx$.
Using the property $\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$, we have $$I = \int_{0}^{\pi/2}\frac{\frac{\pi}{2}-x}{\sin (\frac{\pi}{2}-x)+\cos (\frac{\pi}{2}-x)}dx = \int_{0}^{\pi/2}\frac{\frac{\pi}{2}-x}{\cos x+\sin x}dx$$
Adding the two expressions for $I$, we get $$2I = \int_{0}^{\pi/2}\frac{x}{\sin x+\cos x}dx + \int_{0}^{\pi/2}\frac{\frac{\pi}{2}-x}{\sin x+\cos x}dx = \int_{0}^{\pi/2}\frac{x+\frac{\pi}{2}-x}{\sin x+\cos x}dx$$ $$2I = \int_{0}^{\pi/2}\frac{\frac{\pi}{2}}{\sin x+\cos x}dx = \frac{\pi}{2}\int_{0}^{\pi/2}\frac{1}{\sin x+\cos x}dx$$
$$I = \frac{\pi}{4}\int_{0}^{\pi/2}\frac{1}{\sin x+\cos x}dx$$ We can write $\sin x + \cos x = \sqrt{2}\sin(x+\frac{\pi}{4})$. $$I = \frac{\pi}{4}\int_{0}^{\pi/2}\frac{1}{\sqrt{2}\sin(x+\frac{\pi}{4})}dx = \frac{\pi}{4\sqrt{2}}\int_{0}^{\pi/2}\csc(x+\frac{\pi}{4})dx$$
We know that $\int \csc(x) dx = -\ln|\csc x + \cot x| + C$. Therefore, $$I = \frac{\pi}{4\sqrt{2}}\left[-\ln|\csc(x+\frac{\pi}{4}) + \cot(x+\frac{\pi}{4})|\right]_{0}^{\pi/2}$$ $$I = \frac{\pi}{4\sqrt{2}}\left[-\ln|\csc(\frac{3\pi}{4}) + \cot(\frac{3\pi}{4})| + \ln|\csc(\frac{\pi}{4}) + \cot(\frac{\pi}{4})|\right]$$ $$I = \frac{\pi}{4\sqrt{2}}\left[-\ln|\sqrt{2} - 1| + \ln|\sqrt{2} + 1|\right]$$ $$I = \frac{\pi}{4\sqrt{2}}\ln\left|\frac{\sqrt{2}+1}{\sqrt{2}-1}\right| = \frac{\pi}{4\sqrt{2}}\ln\left|\frac{(\sqrt{2}+1)^2}{2-1}\right| = \frac{\pi}{4\sqrt{2}}\ln(3+2\sqrt{2})$$ $$I = \frac{\pi}{4\sqrt{2}}\ln((\sqrt{2}+1)^2) = \frac{\pi}{4\sqrt{2}}2\ln(\sqrt{2}+1) = \frac{\pi}{2\sqrt{2}}\ln(\sqrt{2}+1)$$
Final Answer: $\frac{\pi}{2\sqrt{2}}\ln(\sqrt{2}+1)$
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