If $$ I(a)=\int_0^\infty\frac{\arctan(ax)-\arctan(x)}{x}\,\mathrm dx, $$ then
In this question the derivative of I should be zero as RHS of a $I$ is constant .
But the answer is given as $A,B,C,D$
Note that $I(a)$ is a function of $a$. Therefore, we can differentiate under the integral to obtain
$$\begin{align} I'(a)&=\frac{d}{da}\int_0^\infty \frac{\arctan(ax)-\arctan(x)}{x}\,dx\\\\ &=\int_0^\infty \frac{\partial}{\partial a}\left(\frac{\arctan(ax)-\arctan(x)}{x}\right)\,dx\\\\ &=\int_0^\infty \frac{1}{1+(ax)^2}\,dx\tag 1\\\\ &=\frac{\pi}{2a}\tag 2 \end{align}$$
Then, using $I(1)=0$, we find after integrating $(2)$ that
$$\bbox[5px,border:2px solid #C0A000]{I(a)=\frac{\pi \log(a)}{2}}\tag 3$$
Then, we see from $(2)$ and $(3)$ that all choices (A), (B), (C), and (D) are true.
NOTE:
The uniform convergence of the integral in $(1)$ for $|a|\ge \delta>0$ (coupled with the convergence of $I(a)$ for all $a\ne0$) guarantees the legitimacy of differentiating under the integral sign.
Note that the conditions of Frullani's Integral are met with $f(x)=\arctan(x)$. So, your integral is just $$\lim_{b\to\infty}(\arctan(b)-0)\ln(a)=\boxed{\frac{\pi}{2}\cdot\ln(a)}$$ Now, consider your options.
