Confusion in options of integration 
If
  $$
I(a)=\int_0^\infty\frac{\arctan(ax)-\arctan(x)}{x}\,\mathrm dx,
$$
  then



*

*(A) $I'(1),I'(2),I'(3)$ are in harmonic progression.

*(B) $I'(2)=\dfrac\pi4$

*(C) $I(\pi)=\dfrac\pi2\log\pi$

*(D) $I'(3)=\dfrac\pi6$  


In this question the derivative of I should be zero as RHS of a $I$ is constant .
But the answer is given as $A,B,C,D$
 A: 
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.

A: 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.
