<s>[INST] integral of $\int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx} $ 
Solve $\int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx} $.  

Answer is 1007.

I tried multiplying $\sqrt{x}-\sqrt{2014-x}\;$,
which results in $\frac{\sqrt{2014-x}(\sqrt{x}-\sqrt{2014-x})}{2x-2014}=$$\frac{\sqrt{2014x-x^2}}{2x-2014}-... \\ =\frac{\sqrt{2014/x-1}}{2-2014/x}-...  
 $
I got stuck so I tried substituting $u=2014-x$,
thus $\int_{0}^{2014}{\frac{u}{\sqrt{2014-u}+u}}du=... ?$
I found the value of 1007 using the value of integrand at x=0, 1007 and 2014.
But cannot solve integral. How can it be solved?
 [/INST] If you substitute $x \mapsto 2014-x$, you get
$$ I = \int_0^{2014} \frac{\sqrt{2014-x}}{\sqrt{x} + \sqrt{2014-x}} \, dx = \int_{0}^{2014} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{2014-x}} \, dx. $$
Hence
\begin{align}
1007 &= \frac{1}{2}\int_0^{2014} dx = \frac{1}{2}\int_0^{2014}\frac{\sqrt{2014-x}}{\sqrt{x} + \sqrt{2014-x}} + \frac{\sqrt{x}}{\sqrt{x} + \sqrt{2014-x}} \, dx\\
&= \frac{1}{2}(I+I) = I.
\end{align}
</s>