$e^{-x}\cdot f(x)=2+\int_{0}^{1}\sqrt{1+t^4}dt$ , for all $x \in (-1,1)$. Find $(f^{-1})^{'}(2)?$

Question

I have tried to find the $\int_{0}^{1}\sqrt{1+t^4}$ but it appears its answer is $\Gamma^2(\frac{1}{4})\cdot \frac{1}{8\cdot\sqrt{\pi}}$. But can anyone tell me how to solve this problem. Here f be a real valued function defined on the interval (−1, 1).

Answer 1

Let $C = 2+\int_{0}^{1}\sqrt{1+t^4}dt$:

$$e^{-x}\cdot f(x)= C$$ $$f(x)= Ce^x$$ $$f^{-1}(x) = \ln\left(\frac{x}{C}\right)$$ $$[f^{-1}(x)]' = \frac{1}{x}$$ $$\therefore [f^{-1}(2)]' = \frac{1}{2}$$