<s>[INST] 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, $A={bh\over 2}$
 [/INST] Let
$$ I=\int^\infty_0 \frac{dx}{(a^2+b^2x^2)(b^2+a^2x^2)}.\tag{1}$$
Changing variable from $x\to\frac1{x}$ gives
$$ I=\int^\infty_0 \frac{x^2dx}{(a^2+b^2x^2)(b^2+a^2x^2)}.\tag{2}$$
So $a^2$(1)+$b^2$(2) gives
$$ a^2I+b^2I=\int^\infty_0 \frac{dx}{b^2+a^2x^2}=\frac{\pi}{2ab}$$
and hence
$$ I=\frac{\pi}{2ab(a^2+b^2)}.$$
So
$$ \frac{2}{\pi}\int^\infty_0 \frac{(ab)^3dx}{(a^2+b^2x^2)(b^2+a^2x^2)} = \frac{(ab)^3}{ab(a^2+b^2)}= \frac{a^2b^2}{a^2+b^2}=\frac{2}{\pi}(ab)^3I=\frac{a^2b^2}{a^2+b^2}=h.$$
</s>