We consider an integral $\int_{0}^{2}{g(x)dx}$, where $g(a)$ is a solution of $x^{5}+x=a$.
Actually, it means that $g^{5}(a)+g(a)-a=0$. Moreover, it somehow possible to reestablish $g(x)$ on $[0, 2]$ by solving the previous equation for various $a$. But this does not lead to the integral's evaluation.
Any piece of advice would be much appreciated.
If $f$ is a continuous non-decreasing function over the interval $[a,b]$, then: $$ \int_{a}^{b}f(x)\,dx + \int_{f(a)}^{f(b)}f^{-1}(x)\,dx = b\, f(b)-a\, f(a).\tag{1} $$ Proof: just draw a picture (or look at here).
By taking $f(x)=x^5+x$ and $[a,b]=[0,1]$ it follows that: $$ \int_{0}^{1}(x^5+x)\,dx + \int_{0}^{2}g(x)\,dx = 2\tag{2} $$ hence: $$ \int_{0}^{2}g(x)\,dx = 2-\left(\frac{1}{6}+\frac{1}{2}\right)=\color{red}{\frac{4}{3}}.\tag{3}$$