Find $\int_0^1 \frac{f(x)}{\sqrt{1+x^2}}dx$ 
Let $f(x)$ be continuous on $[0;1]$, with $f(0) = 0; f(1) = 1$ and
$$\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx = \dfrac{1}{\ln\left(1+\sqrt{2}\right)}$$
Find ${\displaystyle \int_0^1} \dfrac{f(x)}{\sqrt{1+x^2}} \,dx$


*

*Attempt:

I tried to use Cauchy-Scharwz as below:
$$\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx \cdot \int_0^1 \dfrac{f(x)}{\sqrt{1+x^2}} \,dx \geq \left(\int_0^1 \sqrt{[f'(x)]^2 \sqrt{1+x^2}} \cdot \sqrt{\dfrac{f(x)}{\sqrt{1+x^2}}} \,dx\right)^2$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left(\int_0^1 f'(x) \sqrt{f(x)} \,dx\right)^2$$
I was able to find ${\displaystyle \int_0^1} f'(x) \sqrt{f(x)} \,dx = \dfrac{2}{3}$, but the problem is I can't show if the equality is happen or not, so my attempt isn't helpful at all.
Is there a better way to approach this?
 A: Rewriting the given condition as
$$I = \color{blue}{\ln(1+\sqrt 2)}\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx = 1$$
we can notice that
$$I = \color{blue}{\int_0^1 \frac{1}{\sqrt{1+x^2}}dx}\cdot\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx = 1 \tag{1}$$
However, by CS inequality we have
$$\begin{align}
I&=\int_0^1 \frac{1}{\sqrt{1+x^2}}dx\cdot\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx \\ 
&\ge \left(\int f'(x) dx\right)^2 \\
&= (f(1)-f(0))^2 \\
&= 1 \tag{2}\end{align}$$
Hence the equality of CS inequality holds for $$g(x) = \frac{1}{\sqrt{1+x^2}} \ \ \text{ and } \ \ h(x) = [f'(x)]^2 \sqrt{1+x^2}$$
As Martin R said, now solve $h(x) = Cg(x)$ for $f(x)$, then find the integral that you want.
A: From the @VIVID answer, we can then find $f(x)$ and solve for the original integral
As he shows:
$$1 = \int_0^1 \frac{1}{\sqrt{1+x^2}}dx\cdot\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx \geq \left(\int_0^1 f'(x) dx\right)^2 = 1$$
The equality happens when:
$$[f'(x)]^2 \sqrt{1+x^2} = \frac{k}{\sqrt{1+x^2}} \,\,\text{(k is a constant)}$$
$$\implies f'(x) = \sqrt{k} \cdot \frac{1}{\sqrt{1+x^2}} \,\,\,\text{ or }\,\,\, f(x) = \sqrt{k} \,\ln(|\sqrt{x^2 + 1} + x|) + C$$
From the conditions $f(0) = 0; f(1) = 1$, you can find $k$ and $C$. Hence the function will be:
$$f(x) = \frac{1}{\ln(1 + \sqrt{2})} \cdot \ln(|\sqrt{x^2 + 1} + x|)$$
Now solve for the integral:
$$\int_0^1 \frac{f(x)}{\sqrt{1+x^2}} \, dx = \frac{1}{\ln(1 + \sqrt{2})} \cdot \int_0^1 \frac{\ln(|\sqrt{x^2 + 1} + x|)}{\sqrt{1+x^2}} \, dx$$
Let $u = \ln(|\sqrt{x^2 + 1} + x|)$ then the integral become:
$$\frac{1}{\ln(1 + \sqrt{2})} \cdot \int_0^{\ln(1 + \sqrt{2})} u \,du = \color{red}{\frac{1}{2} \cdot \ln(1 + \sqrt{2})}$$
A: VIVID demonstrated that equality holds in the Cauchy-Schwarz inequality
$$ \tag{*}
 1 = \int_0^1 \frac{dx}{\sqrt{1+x^2}} \int_0^1 f'(x)^2 \sqrt{1+x^2} \, dx \ge \left( \int_0^1 f'(x) \right)^2 = 1 \, .
$$
It follows that
$$
 f'(x) = \frac{k}{\sqrt{1+x^2}}
$$
for some (positive) constant $k$. Substituting this back into $(*)$ gives
$$
 1 = k^2 \left(\int_0^1 \frac{dx}{\sqrt{1+x^2}} \right)^2 = k^2 (\ln(1+\sqrt 2))^2
$$
so that $\frac 1k = \ln(1+\sqrt 2)$. Then
$$
 \int_0^1 \frac{f(x)}{\sqrt{1+x^2}} \, dx = \frac 1k \int_0^1 f(x) f'(x) \, dx
= \frac 1{2k}( f^2(1) - f^2(0)) = \frac 1{2k} = \frac 12 \ln(1+\sqrt 2) \,.
$$
