Showing$ \int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$ Showing $$\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$$
We can show this by re-writing $I$ as
$$
\implies I=6\int_{0}^{\infty}\frac{\frac{1-\cos(2x)}{2x}-\frac{1-\cos(3x)}{3x}}{x}\,\mathrm dx,
$$
which is Frullani Integral.
$$J=\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x} dx=[f(\infty)-f(0)]\ln(a/b).$$ Here, $f(x)=\frac{1-\cos(x)}{x},$ hence $I=0.$
So the question is how to show (1), otherwise?

Alternatively, integrate by parts
\begin{align}
&\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx\\
= &\int_{0}^{\infty} \frac{6\sin 2x-6\sin 3x}{x} dx\\
=& \>6 \int_{0}^{\infty} \frac{\sin 2x}{2x} d(2x) 
 - 6 \int_{0}^{\infty} \frac{\sin 3x}{3x} d(3x) \\
 =&\>0
\end{align}
