Let $\int_0^2 f\left(x\right) dx = a+\frac{b}{\log 2}$. Find $a,b$ Let $f$ be a real-valued continuous function on $\mathbb{R}$ such that $2^{f\left(x\right)}+f\left(x\right)=x+1$ for all $x\in \mathbb{R}$. Assume that $\int_0^2 f\left(x\right) dx = a+\dfrac{b}{\log 2}$ with $a,b$ are rational numbers . Find $a,b$.

I have no idea how to use the assumption $2^{f\left(x\right)}+f\left(x\right)=x+1$ for all $x\in \mathbb{R}$ except taking integral of the both sides:
$$\int_0^2 2^{f\left(x\right)}+f\left(x\right) dx=4 $$
Please help me an idea. Thank you.
 A: The function $y+2^y$ is one-to-one (because of its strict monotonicity). Notice that
$$2^0 + 0 = 0 + 1$$
$$2^1 + 1 = 2 + 1$$
just by guessing, and the one-to-one property makes these solutions unique. Thus geometrically we have that:
$$\int_0^2 f(x)\:dx + \int_0^1f^{-1}(y)\:dy = 2\cdot 1 - 0\cdot 0 = 2$$
a rectangle of area $2$ minus a rectangle of area $0$ (with $x=f^{-1}(y) = 2^y+y-1$). Therefore the desired integral is given by
$$\int_0^2f(x)\:dx = 2-\int_0^12^y+y-1\:dy = 2-\left(\frac{2-1}{\log 2}+\frac{1}{2}-1\right) = \frac{5}{2}-\frac{1}{\log 2}$$
A: Let $\alpha = \log 2$, differenital $x + 1 = 2^f + f$ on both sides, we get
$$\begin{align}
1 &= (\alpha 2^f + 1)f'
= (\alpha(x + 1 - f) + 1)f'\\
&= \left[\alpha\left((x + 1)f - \frac{f^2}{2}\right) + f\right]' - \alpha f\\
\implies \alpha f &= \left[\alpha\left((x + 1)f - \frac{f^2}{2}\right) + f - x\right]'
\end{align}
$$
Integrate both sides for $x$ over $[0,2]$ and notice $f(0) = 0$ and $f(2) = 1$, we get
$$\begin{align}
\alpha\int_0^2 f dx
&= \left[\alpha\left((2 + 1)1 - \frac{1^2}{2}\right) + 1 - 2\right]
- \left[\alpha\left((0 + 1)0 - \frac{0^2}{2}\right) + 0 - 0\right]\\
&= \alpha\frac52 - 1
\end{align}
$$
This means $\displaystyle\;\int_0^2 fdx = \frac52 - \frac1{\log 2}$
A: My solution is here :
I hope this will help you!
(https://i.stack.imgur.com/uCEc3.jpg)
