<s>[INST] 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.
 [/INST] 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 $\displaystyle\;\int_0^2 fdx = \frac52 - \frac1{\log 2}$
</s>