Find the exact value of integration $ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$ Can you help me find the exact value for integration with the given steps?
$$ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$$
Some of my attempts as indefinite Integral
$$
\int \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}} \, dx\approx \left(\sqrt{x+1}+\left(-\frac{1}{\sqrt{x+1}+1}-1\right) \sqrt{1-x}+\frac{1}{\sqrt{x+1}+1}-\frac{2 \left(0.707107 \sqrt{x+1}\right)}{\sin }\right)+C
$$ 
Is it considered Improper Integral?
 A: Hint: 
$$
\begin{aligned}
& \int\frac{dx}{\sqrt{1-x}+\sqrt{x+1}+2}\\
& \stackrel{x\to\cos2\phi}= 
\int\frac{\sin{2\phi}\,d\phi}{1+\frac1{\sqrt2}(\sin\phi+\cos\phi)}
=\int\frac{\sin{2\phi}\,d\phi}{1+\sin(\phi+\frac\pi4)}\\
&\stackrel{\phi\to\theta+\frac\pi4}
=\int\frac{\cos{2\theta}\,d\theta}{1+\cos\theta}=\int\frac{2\cos^2{\theta}-1}{1+\cos\theta}\,d\theta\\
&\stackrel{\theta\to2\arctan t}=\int\left[2\left(1-\frac{2}{t^2+1}\right)^2-1\right]\,dt.
\end{aligned}
$$
The rest should not be complicated.
A: Setting $x=\cos(2t)$, we have: 
$1-x=2\sin^2t\;\;$ and $\;\;1+x=2\cos^2t$
A: Substitute $x = \sin 2t $ to have
$$\sqrt{1-x} = \cos t - \sin t, \>\>\>\>\>\sqrt{1+x} = \cos t + \sin t$$
and,
$$\begin{align}
& \int_0^1 \frac{dx}{\sqrt{1-x}+\sqrt{x+1}+2} \\
& = \int_0^{\pi/4}\frac{\cos2t}{1+\cos t}dt \\
&= \int_0^{\pi/4}\frac{2(1+\cos t)^2 -4 (1+\cos t) +1}{1+\cos t}dt \\
& =\int_0^{\pi/4} \left(-2 + 2\cos t + \frac12\sec^2 \frac t2\right)dt \\
&= \left(-2t + 2\sin t + \tan \frac t2\right)\bigg|_0^{\pi/4}= 2\sqrt2-1-\frac\pi2
\end{align}$$
