1. **State the problem:** Evaluate the integral $$\int_{0}^{\pi/2} \frac{1}{(\sqrt{\sin x}+\sqrt{\cos x})^2} \, dx.$$\n\n2. **Rewrite the integrand:** Note that $$(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x + \cos x + 2\sqrt{\sin x \cos x}.$$\nSo the integrand is $$\frac{1}{\sin x + \cos x + 2\sqrt{\sin x \cos x}}.$$\n\n3. **Use the identity for the denominator:** Recall that $$\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$$ and $$\sqrt{\sin x \cos x} = \sqrt{\frac{\sin 2x}{2}} = \frac{\sqrt{\sin 2x}}{\sqrt{2}}.$$\n\n4. **Simplify the denominator:**\n$$\sin x + \cos x + 2\sqrt{\sin x \cos x} = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) + 2 \cdot \frac{\sqrt{\sin 2x}}{\sqrt{2}} = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) + \sqrt{2} \sqrt{\sin 2x}.$$\n\n5. **Rewrite the integrand:**\n$$\frac{1}{(\sqrt{\sin x}+\sqrt{\cos x})^2} = \frac{1}{\sqrt{2} \left( \sin\left(x + \frac{\pi}{4}\right) + \sqrt{\sin 2x} \right)} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sin\left(x + \frac{\pi}{4}\right) + \sqrt{\sin 2x}}.$$\n\n6. **Use substitution:** Let $$t = x + \frac{\pi}{4}.$$ Then when $$x=0, t=\frac{\pi}{4}$$ and when $$x=\frac{\pi}{2}, t=\frac{3\pi}{4}.$$ Also, $$\sin 2x = \sin(2t - \frac{\pi}{2}) = \cos 2t.$$\n\n7. **Rewrite the integral in terms of $t$:**\n$$I = \frac{1}{\sqrt{2}} \int_{\pi/4}^{3\pi/4} \frac{1}{\sin t + \sqrt{\cos 2t}} \, dt.$$\n\n8. **Simplify $\sqrt{\cos 2t}$:** For $t \in [\pi/4, 3\pi/4]$, $\cos 2t \leq 0$, so $\sqrt{\cos 2t}$ is not real. Instead, rewrite $\sqrt{\sin 2x}$ as $\sqrt{\sin 2x} = \sqrt{2 \sin x \cos x}$, but this approach is complicated.\n\n9. **Alternative approach:** Use the identity\n$$(\sqrt{\sin x} + \sqrt{\cos x})^2 + (\sqrt{\sin x} - \sqrt{\cos x})^2 = 2(\sin x + \cos x) = 2.$$\n\n10. **Express the integrand in terms of the other expression:**\n$$\frac{1}{(\sqrt{\sin x} + \sqrt{\cos x})^2} = \frac{(\sqrt{\sin x} - \sqrt{\cos x})^2}{2} = \frac{\sin x + \cos x - 2\sqrt{\sin x \cos x}}{2}.$$\n\n11. **Rewrite the integral:**\n$$I = \int_0^{\pi/2} \frac{\sin x + \cos x - 2\sqrt{\sin x \cos x}}{2} \, dx = \frac{1}{2} \int_0^{\pi/2} (\sin x + \cos x) \, dx - \int_0^{\pi/2} \sqrt{\sin x \cos x} \, dx.$$\n\n12. **Calculate the first integral:**\n$$\int_0^{\pi/2} (\sin x + \cos x) \, dx = \left[-\cos x + \sin x \right]_0^{\pi/2} = (-\cos \frac{\pi}{2} + \sin \frac{\pi}{2}) - (-\cos 0 + \sin 0) = (0 + 1) - (-1 + 0) = 1 + 1 = 2.$$\n\n13. **Calculate the second integral:**\n$$\int_0^{\pi/2} \sqrt{\sin x \cos x} \, dx = \int_0^{\pi/2} \sqrt{\frac{\sin 2x}{2}} \, dx = \frac{1}{\sqrt{2}} \int_0^{\pi/2} \sqrt{\sin 2x} \, dx.$$\n\n14. **Substitute $u = 2x$, $du = 2 dx$, $dx = \frac{du}{2}$:**\n$$\int_0^{\pi/2} \sqrt{\sin 2x} \, dx = \int_0^{\pi} \sqrt{\sin u} \cdot \frac{du}{2} = \frac{1}{2} \int_0^{\pi} \sqrt{\sin u} \, du.$$\n\n15. **Use Beta function or known integral:**\n$$\int_0^{\pi} \sin^m u \, du = \sqrt{\pi} \frac{\Gamma\left(\frac{m+1}{2}\right)}{\Gamma\left(\frac{m}{2} + 1\right)}$$ for $m > -1$. For $m=\frac{1}{2}$,\n$$\int_0^{\pi} \sqrt{\sin u} \, du = \sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{5}{4})}.$$\n\n16. **Therefore:**\n$$\int_0^{\pi/2} \sqrt{\sin x \cos x} \, dx = \frac{1}{2\sqrt{2}} \int_0^{\pi} \sqrt{\sin u} \, du = \frac{1}{2\sqrt{2}} \sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{5}{4})}.$$\n\n17. **Final expression for $I$:**\n$$I = \frac{1}{2} \times 2 - \frac{1}{2\sqrt{2}} \sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{5}{4})} = 1 - \frac{\sqrt{\pi}}{2\sqrt{2}} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{5}{4})}.$$\n\n**Answer:** $$\boxed{1 - \frac{\sqrt{\pi}}{2\sqrt{2}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)}}.$$