Question

The value of the integral ${\int }_0^8\frac{x^2}{x^2-8x+32}dx$ is equal to

Answer 1

Answer: 8

Let, $I={\int }_0^8\frac{x^2}{x\left(x-8\right)+32}dx$
So, $I={\int }_0^8\frac{{\left(8-x\right)}^2}{\left(-x\right)\left(8-x\right)+32}dx\left\{{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right\}$
Adding both equations, we get,
$2I={\int }_0^8\frac{2\left(x^2-8x+32\right)}{x^2-8x+32}dx=16$
$\Rightarrow I=8$