Question

The value of the integral $\displaystyle \int _{0}^{8}\frac{x^{2}}{x^{2} - 8 x + 32}dx$ is equal to

Answer 1

Let, $I=\displaystyle \int _{0}^{8} \frac{x^{2}}{x \left(x - 8\right) + 32} d x$
So, $I=\displaystyle \int _{0}^{8}\frac{\left(8 - x\right)^{2}}{\left(- x\right) \left(8 - x\right) + 32}dx\left\{\displaystyle \int _{a}^{b} f \left(x\right) d x = \displaystyle \int _{a}^{b} f \left(a + b - x\right) d x\right\}$
Adding both equations, we get,
$2I=\displaystyle \int _{0}^{8}\frac{2 \left(x^{2} - 8 x + 32\right)}{x^{2} - 8 x + 32} d x = 16$
$\Rightarrow I=8$