Question

The value of the integral $\displaystyle \int _{0}^{4}\frac{x^{2}}{x^{2} - 4 x + 8}dx$ is equal to

Answer 1

Let $I=\displaystyle \int _{0}^{4} \frac{x^{2}}{x \left(x - 4\right) + 8}dx$
Using $\displaystyle \int _{a}^{b}f\left(x\right)dx=\displaystyle \int _{a}^{b}f\left(a + b - x\right)dx$ , we get,
$I=\displaystyle \int _{0}^{4}\frac{\left(4 - x\right)^{2}}{\left(4 - x\right) \left(- x\right) + 8}dx$
Adding both the equations, we get,
$2I=\displaystyle \int _{0}^{4}\frac{2 \left(x^{2} - 4 x + 8\right)}{x^{2} - 4 x + 8} d x = 8$
$\Rightarrow I=4$