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?

Don't need to do that substitution again and again just apply this formula 
$$\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$$
$$I=\int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx}$$
Apply the formula
$$\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$$
$$  I=\int_{0}^{2014}{\frac{\sqrt{2014-(2014+0-x)}}{\sqrt{2014+0-x}+\sqrt{2014-(2014+0-x)}}dx}=\int_{0}^{2014}{\frac{\sqrt x}{\sqrt{2014-x}+\sqrt{x}}dx}$$
add both integrals 
$$2I=\int_{0}^{2014}dx$$
$$2I=2014$$
$$I=1007$$
