calculus - Prove $int^infty_0 frac{frac{1}{1+(bx)^2}-frac{1}{1+(ax)^2}}{x}dx = ln(frac{a}{b})$ with Frullani Integrals

Prove $$\int^\infty_0 \frac{\frac{1}{1+(bx)^2}-\frac{1}{1+(ax)^2}}{x}dx = \ln(\frac{a}{b})$$

I'm supposed to use Frulanni integrals and use the fact that $\int^\infty_0 \frac{f(bx)-f(ax)}{x}dx$ since this equals $[f(\infty)-f(0)] \ln(\frac{b}{a})$

Unfortunately, I can't figure out how to do such a transformation. Any help would be appreciated!

UPDATE: Here's what I tried.

Let $f(t)=\frac{1}{1+t^2}$ and $b^2=\frac{1}{d}$ and $a^2=\frac{1}{c}$

Then we have $$\int^\infty_0 \frac{f(td)-f(tc)}{x}=[f(\infty)-f(0)]\ln(\frac{d}{c})=(1-1)\ln(\frac{d}{c})=0.$$ but I don't think we're allowed to get 0 so I must have done something wrong.

Answer

So, let $f(t)=1/(1+t^2)$. Then, your integral can be written
$$\int_0^{+\infty}\frac{f(bx)-f(ax)}{x}\,dx$$
and, as you point out, it is a Frullani integral, with value
$$(f(+\infty)-f(0))\ln(b/a)$$
Now

$$\lim_{t\to+\infty}f(t)=0$$
and
$$f(0)=1.$$
Hence, the value of your integral is
$$-\ln(b/a)=-(\ln b-\ln a)=\ln a-\ln b=\ln(a/b)$$
(I made that calculation in detail since I think that was your problem) as desired.