$\text{ Let }g\text{ be a differentiable function satisfies }{\int }_0^x (x-t+1)g\left(t\right)dt=x^4+x^2\mathrm{\forall }x\ge 0$, $\text{ then the value of }{\int }_0^1 \frac{12}{g^1\left(x\right)+g\left(x\right)+10}dx=$

 

$\begin{array}{l}\mathrm{x}{\int }_0^{\mathrm{x}} \mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}+{\int }_0^x (1-\mathrm{t})\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}={\mathrm{x}}^4+{\mathrm{x}}^2\\ \text{ differentiate with respect to }\mathrm{x}\\ {\int }_0^{\mathrm{x}} \mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}+\mathrm{xg}\left(\mathrm{x}\right)+\left(1-x\right)g\left(x\right)=4{\mathrm{x}}^3+2\mathrm{x}\end{array}$

$\begin{array}{l}\text{ again differentiate with respective to }\mathrm{x}\\ \Rightarrow g^{\mathrm{\prime }}\left(x\right)+g\left(x\right)=12x^2+2\\ \therefore {\int }_0^1 \frac{12}{g^1\left(x\right)+g\left(x\right)+10}dx={\int }_0^1 \frac{12}{12\left({\mathrm{x}}^2+1\right)}\mathrm{dx}={\left[{\tan }^{-1}x\right]}_0^1=\frac{\pi }{4}\end{array}$