Let $q\in(0,1)$ and $x\in\{0,1\}$. The likelihood of a Bernoulli trial is $$q^x(1-q)^{(1-x)}.$$ If we take an integration with respect to $q$ over $[0,1]$, the value should be $\frac{1}{2}$: $$\int^1_0q^x(1-q)^{(1-x)}dq=B(2,1)=B(1,2)=\frac{1}{2}.$$ My question is, how can we derive the integrated value over $[a,b]$, for some $0\leq a\leq b\leq 1$?
Would there be closed form solution for the following integral, not using the incomplete beta function? $$\int^b_aq^x(1-q)^{(1-x)}dq?$$
Not really. You won't be able to get away without using special functions.
You can, however, use the hypergeometric function $_2F_1$, but it's really just the incomplete beta in disguise. We set out to evaluate the integral $$f=\int_0^1 t^{b-1}(1-t)^{c-b-1}(1-xt)^{-a}dt.\qquad \text{Re }c>\text{Re }b>0$$ We have that $$(1-u)^{-s}=\,_1F_0(s;;u)=\sum_{n\ge0}\frac{(s)_n}{n!}u^n$$ where $(x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}$. So then we get $$\begin{align} f&=\int_0^1 t^{b-1}(1-t)^{c-b-1}\sum_{n\ge0}\frac{(a)_n}{n!}x^nt^ndt\\ &=\sum_{n\ge0}\frac{(a)_n}{n!}x^n\int_0^1 t^{n+b-1}(1-t)^{c-b-1}dt\\ &=\sum_{n\ge0}\frac{(a)_n}{n!}x^n\frac{\Gamma(n+b)\Gamma(c-b)}{\Gamma(n+c)}\\ &=\frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}\sum_{n\ge0}\frac{(a)_n}{n!}x^n\frac{\Gamma(n+b)\Gamma(c)}{\Gamma(b)\Gamma(n+c)}\\ &=\mathrm{B}(b,c-b)\sum_{n\ge0}\frac{(a)_n(b)_n}{(c)_n}\frac{x^n}{n!}\\ &=\mathrm{B}(b,c-b)\,_2F_1(a,b;c;x). \end{align}$$ Then we see that $$\begin{align} \mathrm{B}(x;u,v)&=\int_0^x t^{u-1}(1-t)^{v-1}dt\\ &=x^u\int_0^1 t^{u-1}(1-xt)^{v-1}dt\\ &=x^u\mathrm{B}(u,1)\,_2F_1(1-v,u;1+u;x)\\ &=\frac{x^u}{u}\,_2F_1(u,1-v;1+u;x). \end{align}$$ Your integral is given by $$\int_a^b q^{x}(1-q)^{1-x}dq=\frac{z^{x+1}}{x+1}\,_2F_1(x+1,x-1;x+2;z)\bigg|_{z=a}^{z=b}.$$