The value of the integral ${\int }_0^4\frac{x^2}{x^2-4x+8}dx$ is equal to

Let $I={\int }_0^4\frac{x^2}{x\left(x-4\right)+8}dx$

Using $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx$, we get, 
 $I={\int }_0^4\frac{{\left(4-x\right)}^2}{\left(4-x\right)\left(-x\right)+8}dx$
Adding both the equations, we get,

 $2I=\underset{0}{\overset{4}{\int }}\frac{2\left(x^2-4x+8\right)}{x^2-4x+8}dx=8$

$\Rightarrow I=4$