Question

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

Answer 1

Answer: 4

Let $I={\int }_0^4\frac{x^2}{x\left(x-4\right)+8}dx$
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$ , 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,
$2I={\int }_0^4\frac{2\left(x^2-4x+8\right)}{x^2-4x+8}dx=8$
$\Rightarrow I=4$