Prove the following integrals' equality

Question

$$ \int_a^b \frac{x^2 - 2ax + a^2}{2x^2 - 2(a + b)x + a^2 + b^2} dx = \int_a^b \frac{x^2 - 2bx + b^2}{2x^2 - 2(a + b)x + a^2 + b^2} dx $$

I tried to solve this by Partial Fractions, and did $$ 2x^2 - 2(a+b)x + a^2+b^2 = 2x(x-(a+b)) + (a^2+b^2) $$ but I can't see something else to do with this. Could you give me any hint? I'm stuck.

Answer 1

Hint as requested: note that the left hand side integral is $$L=\int_a^b \frac{(x-a)^2}{(x-a)^2+(x-b)^2}\,dx$$ and then substitute $x=a+b-y$.

Answer 2

We make the substitution $u = a+b-x$. Then we have \begin{align*}\int_{a}^{b}\frac{x^2-2ax+a^2}{2x^2-2(a+b)x+a^2+b^2} \mathop{}\!\mathrm{d}x &= \int_{a}^{b}\frac{(x-a)^2}{(x-a)^2+(x-b)^2}\mathop{}\!\mathrm{d}x \\ &= \int_{b}^{a}\frac{(b-u)^2}{(b-u)^2+(a-u)^2}\left(-\mathop{}\!\mathrm{d}u\right) \\ &= \int_{a}^{b}\frac{(u-b)^2}{(u-a)^2+(u-b)^2}\mathop{}\!\mathrm{d}u \\ &= \int_{a}^{b}\frac{(x-b)^2}{(x-a)^2+(x-b)^2}\mathop{}\!\mathrm{d}x \\ &= \int_{a}^{b}\frac{x^2-2bx+b^2}{2x^2-2(a+b)x+a^2+b^2}\end{align*} as required.