Does it need substitution?

$\large \sqrt{1-x} + \sqrt{1+x}$

$\large = \sqrt{ \left ( \sqrt{1-x} + \sqrt{1+x} \right )^2 }$

$\large = \left ( 2 + 2 \sqrt{1-x^2} \right )^{ \frac {1}{2} }$

The integral becomes

$\displaystyle \frac {1}{2} \int_0^1 \ln \left ( 2 + 2 \sqrt{1-x^2} \right ) \mathrm{d}x$

Use properties of logarithm $\ln(AB) = \ln(A) + \ln(B)$ for $A, B>0$

$\displaystyle = \frac {1}{2} \int_0^1 \ln 2 \space \mathrm{d}x + \frac {1}{2} \int_0^1 \ln \left (1 + \sqrt{1-x^2} \right ) \mathrm{d}x$

$\displaystyle = \frac {1}{2} \ln 2 + \frac {1}{2} \int_0^1 \ln \left (1 + \sqrt{1-x^2} \right ) \mathrm{d}x$

Let $x = \sin \theta \Rightarrow \mathrm{d}x = \cos \theta \space \mathrm{d} \theta$

$\displaystyle = \frac {1}{2} \ln 2 + \frac {1}{2} \int_0^{\pi /2} \ln (1 + \cos \theta) \cos \theta \space \mathrm{d} \theta$

Integrate by parts,

Let $u = \ln(1 + \cos \theta) , \mathrm{d}v = \cos \theta \space \mathrm{d} \theta$

$\Rightarrow \mathrm{d}u = - \frac {- \sin \theta}{1 + \cos \theta} \mathrm{d} \theta, v = \sin \theta$

With $\displaystyle \int u \mathrm{d}v = uv - \int v \mathrm{d}u$

Apply the limits and simplify, we get

$\displaystyle = \frac {1}{2} \ln 2 + \frac {1}{2} \int_0^{\pi /2} \frac { \sin^2 \theta }{1 + \cos \theta } \space \mathrm{d} \theta$

$\displaystyle = \frac {1}{2} \ln 2 + \frac {1}{2} \int_0^{\pi /2} \frac { 1 - \cos^2 \theta }{1 + \cos \theta } \space \mathrm{d} \theta$

$\displaystyle = \frac {1}{2} \ln 2 + \frac {1}{2} \int_0^{\pi /2} (1 - \cos \theta ) \space \mathrm{d} \theta$

$\displaystyle = \frac {1}{2} \ln 2 + \frac {1}{2} \left ( \frac { \pi}{2} - 1 \right ) = \LARGE \boxed{ \frac {2\ln 2 + \pi -2}{4} }$

Let $I=\int{log(\sqrt{1-x}+\sqrt{1+x})dx}$

Let $u=log(\sqrt{1-x}+\sqrt{1+x})$ and $dv=dx$ i.e, $v=x$

Then, $du=\frac{1}{\sqrt{1-x}+\sqrt{1+x}}(-\frac{1}{2\sqrt{1-x}}+\frac{1}{2\sqrt{1+x}})dx=\frac{1}{2}\frac{\sqrt{1-x^{2}}-1}{x\sqrt{1-x^{2}}}$ (Simplifying)

Now, Applying Integration by parts,

$I=uv-\int{vdu}$ $=xlog(\sqrt{1-x}+\sqrt{1+x})-\frac{1}{2}\int{x\frac{\sqrt{1-x^{2}}-1}{x\sqrt{1-x^{2}}}}$ $=xlog(\sqrt{1-x}+\sqrt{1+x})-\frac{1}{2}\int{dx}+\frac{1}{2}\int{\frac{1}{\sqrt{1-x^{2}}}dx}=xlog(\sqrt{1-x}+\sqrt{1+x})-\frac{1}{2}x+\frac{1}{2}sin^{-1}x+C$ where C is constant

Hence, $\int_{0}^{1}{log(\sqrt{1-x}+\sqrt{1+x})dx}$ $=log\sqrt{2}-\frac{1}{2}+\frac{1}{2}sin^{-1}1-0.log2+\frac{1}{2}\times0-\frac{1}{2}sin^{-1}0$ $=log\sqrt{2}-\frac{1}{2}+\frac{π}{4}$ $=0.63197175=0.632$

So the answer to the question is SUBSTITUTION IS NOT NECESSARY

Clerarly Speaking I used Substitution

$\displaystyle\int$ (log( $\sqrt(1-x)$ + $\sqrt(1+x)$ ) $dx$

after applying substitution and by part integration many times we get (sorry I am too lazy can't handle too much laTeX here so directly going to the last expression)

$\frac{1}{2}$ (- $x$ - $log(-x)$ + $log(1-(\sqrt(x+1)$ ))- $log((\sqrt(1-x)$ - $\sqrt(1+x)$ + $2$ ) - $log(\sqrt(x+1))+1)+2(x+1)(log(\sqrt1-x$ + $\sqrt(1+x)+(log(\sqrt(1-x))+(\sqrt(1+x))+2)+2(sin^{-1}$ . $\frac{\sqrt(x+1)}{(\sqrt2)}$ -1)

On limition It from $0$ to $1$ we get $\boxed{631972....}$

$\begin{aligned} I & = \int_0^1 \ln (\sqrt{1-x} + \sqrt{1+x}) \ dx \quad \quad \small \color{#3D99F6}{\text{By integration by parts}} \\ & = x \ln (\sqrt{1-x} + \sqrt{1+x}) \bigg|_0^1 - \int_0^1 \frac {\frac x2\left(\frac 1{\sqrt{1+x}} - \frac 1{\sqrt{1-x}} \right)}{\sqrt{1-x} + \sqrt{1+x}} \ dx \\ & = \frac {\ln 2}2 + \int_0^1 \frac {(\sqrt{1+x} - \sqrt{1-x})^2}{4\sqrt{1-x^2}} \ dx \\ & = \frac {\ln 2}2 + \int_0^1 \frac {2- 2 \sqrt{1-x^2}}{4\sqrt{1-x^2}} \ dx \\ & = \frac {\ln 2}2 + \frac 12 \int_0^1 \frac 1{\sqrt{1-x^2}} \ dx - \frac 12 \int_0^1 \ dx \\ & = \frac {\ln 2}2 + \frac 12 \sin^{-1} x \bigg|_0^1 - \frac 12 x \bigg|_0^1 \\ & = \frac {\ln 2}2 + \frac \pi 4 - \frac 12 \\ & \approx \boxed{0.632} \end{aligned}$

I Used the substituion x = cos(2y) and used integration by parts

Let $p=\sqrt{1+x}+\sqrt{1-x}$ , so we have $x=\frac{1}{2}\sqrt{4p^2-p^4}$ and

$\large\mathrm{d}x=\frac{8p-4p^3}{4\sqrt{4p^2-p^4}}\mathrm{d}p$

The integral becomes

$\displaystyle I=-\int_{\sqrt{2}}^2 \frac{8p-4p^3}{4\sqrt{4p^2-p^4}} \log p \mathrm{d}p = -\frac{1}{2}\int_{\sqrt{2}}^2\log p \mathrm{d}\left(\sqrt{4p^2-p^4}\right)$

Using integral by parts:

Let $u=\log p, dv=d\left(\sqrt{4p^2-p^4}\right)$

$\Rightarrow \mathrm{d}u=\frac{1}{p}, v=\sqrt{4p^2-p^4}=p\sqrt{4-p^2}$

We get,

$\begin{aligned}I&=-\frac{1}{2} \left(\sqrt{4p^2-p^4} \log p -\displaystyle \int_{\sqrt{2}}^2\sqrt{4-p^2} \mathrm{d}p \right)\\ &=-\frac{1}{2} \left(\sqrt{4p^2-p^4} \log p-\left(\frac{p\sqrt{4-p^2}}{2}+2\sin^{-1}\frac{p}{2}\right)\right) \\ &=\frac{1}{2}\sqrt{4p^2-p^4}\left(\frac{1}{2}-\log p\right)+\sin^{-1}\frac{p}{2}\end{aligned}$

Now, apply the limits:

$\large I=\frac{\pi}{2}-\left(\frac{1}{2}-\frac{1}{2}\log2+\frac{\pi}{4}\right)=\frac{\pi-2+\log4}{4}=\boxed{0.632}$

Let's do it without substitution.It's very easy.It just takes 3 steps. Take ln(sqrt(1−x)+sqrt(1+x)) as first function and 1 as second function.As we know that it breaks the integral into two integrals,the first one is already calculated as integration of 1 is x and for the second integral rationalize the denominator and it will automatically convert into a very simple integral. It just takes 3 steps.

Another substitution is: x= cost; we get ln(sqrt(2)[ sin t/2 + cost/2]) after using:f(x )=f(a+b-x) we get: ln(2cos t/2)cos t dt after IBP: (1/2)ln(2) + [ integral( 0 to pi/2(sin^2x/2)dx] which equals: $\dfrac { \ln{2} }{ 2 } +\frac { \pi }{ 4 } -\frac { 1 }{ 2 }$