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, $A={bh\over 2}$
 A: Hint: Observe if you have a right  triangle with legs $a, b$ and $h$ is as specified, then we see that
\begin{align}
\text{Area} = \frac{a\cdot b}{2} = \frac{h\cdot \sqrt{a^2+b^2}}{2} \ \ \Rightarrow \ \ h^2 = \frac{a^2b^2}{a^2+b^2}.
\end{align}
One can show that the integral
\begin{align}
\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} = h^2.
\end{align}
A: 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
$$ \frac{2}{\pi}\int^\infty_0 \frac{(ab)^3dx}{(a^2+b^2x^2)(b^2+a^2x^2)} = \frac{(ab)^3}{ab(a^2+b^2)}= \frac{a^2b^2}{a^2+b^2}=\frac{2}{\pi}(ab)^3I=\frac{a^2b^2}{a^2+b^2}=h.$$
A: 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.
