# Find $\int_0^1 \dfrac{f(x)}{\sqrt{1+x^2}}$

## 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?

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?