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}=$$\frac{\sqrt{2014x-x^2}}{2x-2014}-... \\ =\frac{\sqrt{2014/x-1}}{2-2014/x}-...  
 $
I got stuck so I tried substituting $u=2014-x$,
thus $\int_{0}^{2014}{\frac{u}{\sqrt{2014-u}+u}}du=... ?$
I found the value of 1007 using the value of integrand at x=0, 1007 and 2014.
But cannot solve integral. How can it be solved?
 A: Let $I = \int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx}$.  Then, as you said, consider the substitution $u = 2014 - x$.
In particular, 
$$I = \int_{2014}^0 \frac{\sqrt{u}}{\sqrt{2014 - u} + \sqrt{u}} \cdot (-1) \,du = \int_0^{2014} \frac{\sqrt{u}}{\sqrt{2014 - u} + \sqrt{u}} \,du$$
Relabel the $u$ as $x$, then, adding the integrals together, we get
$$2I = \int_0^{2014} 1 \,dx$$
which gives you the result.
A: If you substitute $x \mapsto 2014-x$, you get
$$ 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. $$
Hence
\begin{align}
1007 &= \frac{1}{2}\int_0^{2014} dx = \frac{1}{2}\int_0^{2014}\frac{\sqrt{2014-x}}{\sqrt{x} + \sqrt{2014-x}} + \frac{\sqrt{x}}{\sqrt{x} + \sqrt{2014-x}} \, dx\\
&= \frac{1}{2}(I+I) = I.
\end{align}
A: $$let \int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx}=A \\ put \ 2014-x=u \\ \implies \int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx}=\int_{2014}^{0}{\frac{\sqrt{u}}{\sqrt{u}+\sqrt{2014-u}}(-du)}\\=\int_{0}^{2014}{\frac{\sqrt{u}}{\sqrt{u}+\sqrt{2014-u}}(du)}=A \\ add \ both \ integrals \ \implies 2(A)=\int_0^{2014}dx \implies A=1007$$
A: Don't need to do that substitution again and again just apply this formula 
$$\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$$
