Evaluate $\int_0^\infty \frac{\arctan(3x) - \arctan(9x)}{x} {dx}$ 
Evaluate the integral $$\int_0^\infty \frac{\arctan(3x) - \arctan(9x)}{x} {dx}.$$

I tried to split this into 2 integrals and then using the substitution $t = \arctan(3x)$ but I got nowhere.
 A: Method 1 A difference of arctangents is exactly the kind of quantity produced when computing the definite integral of the familiar integrand $\frac{1}{1 + y^2}$, which suggests we rewrite the integral (replacing the numerical constants with general constants $a, b > 0$) as a double integral:
$$\int_0^{\infty} \int_{ax}^{bx} \frac{1}{x(1 + y^2)} dy \,dx.$$
Reversing the order of integration (which requires some justification that is often suppressed when learning multiple integration) gives
$$\int_0^{\infty} \int_{y/b}^{y/a} \frac{1}{x(1 + y^2)} dx \,dy.$$

 Now, integrating the inner integral and then the outer one is straightforward; the result should be $$\frac{\pi}{2} \log \frac{b}{a},$$ which in your case is $$-\frac{\pi}{2}\log 3.$$

Method 2 By request, here's an alternate solution that uses the (perhaps undertaught) method of differentiating under the integral:
Regard the integral $$\int_0^{\infty} \frac{\arctan bx - \arctan ax}{x} \,dx$$ as a function $I(b)$ of $b$. Then, one can justify that when computing the derivative $I'(b)$ we can differentiate under the integral sign:
\begin{align}
I'(b)
    &= \frac{d}{db} \int_0^{\infty} \frac{\arctan bx - \arctan ax}{x} \,dx \\
    &= \int_0^{\infty} \frac{d}{db}\left(\frac{\arctan bx - \arctan ax}{x}\right) \,dx \\
    &= \int_0^{\infty} \frac{dx}{1 + (bx)^2} \\
    &= \left.\frac{1}{b} \arctan bx \,\right\vert_0^{\infty} \\
    &= \frac{\pi}{2b} \textrm{.}
\end{align}
Integrating gives that
$$I(b) = \frac{\pi}{2} \log b + I_0$$
for some constant $I_0$. On the one hand, $I(a) = \frac{\pi}{2} \log a + I_0$, and on the other, $$I(a) = \int_0^{\infty} \frac{\arctan ax - \arctan ax}{x} \,dx = 0,$$
so $I_0 = -\frac{\pi}{2} \log a$, and thus the original integral is
$$\int_0^{\infty} \frac{\arctan bx - \arctan ax}{x} \,dx = I(b) = \frac{\pi}{2} \log b - \frac{\pi}{2} \log a = \frac{\pi}{2} \log \frac{b}{a},$$
which agrees with the solution from the first method.
A: An easy way to evaluate the integral is using Frullani's theorem
$$\int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx=\bigg[f(0)-f(\infty)\bigg]\ln\left(\frac{b}{a}\right)$$
Taking $f(x)=\arctan(x)$ then the integral is simply evaluated to
$$-\frac{\pi}{2}\ln3$$
