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?

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
$$
 \int_0^1 \frac{f(x)}{\sqrt{1+x^2}} \, dx = \frac 1k \int_0^1 f(x) f'(x) \, dx
= \frac 1{2k}( f^2(1) - f^2(0)) = \frac 1{2k} = \frac 12 \ln(1+\sqrt 2) \,.
$$
