Integral related to Pythagoras theorem ${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2$. Integral related to Pythagoras theorem
Triangle ABC is a right angle triangle, where Angle $ABC=90^o$.
$h$ is perpendicular to the hypotenuse AC and meet at angle ABC.
Where $a$ and $b$ are two small sides
How can I Show that h can be represented in term of this integral $(1)$

$${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2\tag1$$.

Any hints on this can be relate to Pythagoras theorem
Basic formulas : $AC^2=AB^2+BC^2$ and area, $A={bh\over 2}$

Considering the integral $$I=\int{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\, dx$$ using partial fraction decomposition, we have $$I=\int\left(\frac{a^5 b^3}{\left(a^4-b^4\right) \left(a^2 x^2+b^2\right)}-\frac{a^3
   b^5}{\left(a^4-b^4\right) \left(a^2+b^2 x^2\right)}\right)\,dx$$ from which $$I=\frac{a^3 b^3 \left(a^2 \tan ^{-1}\left(\frac{a x}{b}\right)-b^2 \tan
   ^{-1}\left(\frac{b x}{a}\right)\right)}{a^5 b-a b^5}$$ Using the given bounds $$J=\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\, dx=\frac{\pi  a^3 b^3}{2 a^3 b+2 a b^3}=\frac{\pi  a^2 b^2}{2 \left(a^2+b^2\right)}$$
$$K={2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\, dx=\frac{a^2 b^2}{a^2+b^2}$$ from which you can easily finish, I am sure.
