Evaluation of $ \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$

By collecting all the suggestions, you should be able to prove that: $$\frac{2t}{(t+1)(t^4+1)}=\frac{t}{1+t^4}+\frac{t^3}{1+t^4}+\frac{1-t^2}{1+t^4}-\frac{1}{1+t},$$ and since $(1+t^4)=(1+\sqrt{2}t+t^2)(1-\sqrt{2}t+t^2)$, it follows that: $$\int\frac{2t\,dt}{(t+1)(t^4+1)}=\frac{1}{2}\arctan t^2+\frac{1}{4}\log(1+t^4)+\frac{1}{2\sqrt{2}}\log\frac{1+\sqrt{2}t+t^2}{1-\sqrt{2}t+t^2}-\log(1+t).$$ $\\$

Addendum

Suggestion for evaluating $\displaystyle{\int \frac{1 - t^2}{t^4 + 1}\,\mathrm{d}t}$: Write $$ \begin{aligned} \int- \frac{t^2 - 1}{t^4 + 1}\,\mathrm{d}t &= -\int \frac{1 - \dfrac{1}{t^2}}{t^2 + \dfrac{1}{t^2}}\,\mathrm{d}t \\ &=-\int \frac{1 - \dfrac{1}{t^2}}{\left(t + \dfrac{1}{t}\right)^2 - 2}\,\mathrm{d}t \end{aligned} $$

Now, set $\displaystyle{u = t + \frac{1}{t}}$ and $\mathrm{d}u = \left(1 - \dfrac{1}{t^2}\right)\,\mathrm{d}t$: $$ \begin{aligned} -\int \frac{\mathrm{d}u}{u^2 - 2} &= \dfrac{1}{2\sqrt{2}}\ln\left|\frac{u + \sqrt{2}}{u - \sqrt{2}} \right| + C \\ &=\frac{1}{2\sqrt{2}}\ln\left|\frac{t^2 + t\sqrt{2} +1}{t^2 - t\sqrt{2}+1} \right| + C \end{aligned} $$ Note that $t^2 + t\sqrt{2}+1 = \left(t + \dfrac{1}{\sqrt{2}}\right)^{\!2} + \dfrac{1}{2} >0$ and $t^2 - t\sqrt{2}+1 = \left(t - \dfrac{1}{\sqrt{2}}\right)^{\!2} + \dfrac{1}{2} >0$, then we can get rid of the absolute value bars: $$\int \frac{1-t^2}{t^4 + 1}\,\mathrm{d}t = \frac{1}{2\sqrt{2}}\ln\!\left(\dfrac{t^2 + t\sqrt{2}+1}{t^2 - t\sqrt{2}+1}\right) + C $$

This is a very famous technique.