Your professor is right. Note that
$$ \tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}, \tanh(2x)=\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}=\frac{(e^x-e^{-x})(e^x+e^{-x})}{e^{2x}+e^{-2x}}$$
and hence
\begin{eqnarray*}
\int_0^\infty\frac{\tanh(x)\tanh(2x)}{x^2}dx&=&\int_0^\infty\frac{(e^{x}-e^{-x})^2}{x^2(e^{2x}+e^{-2x})}dx\\
&=&\int_0^\infty\frac{e^{2x}-2+e^{-2x}}{x^2(e^{2x}+e^{-2x})}dx.
\end{eqnarray*}
Now define
$$ I(a)=\int_0^\infty\frac{e^{ax}-2+e^{-ax}}{x^2(e^{2x}+e^{-2x})}dx$$
to get
\begin{eqnarray}
I''(a)&=&\int_0^\infty\frac{e^{(-a-2)x}+e^{(-a+2)x}}{1+e^{-4x}}dx,\\
&=&\int_0^\infty\sum_{n=0}^\infty(-1)^n(e^{(-a-2)x}+e^{(-a+2)x})e^{-4nx}dx\\
&=&\sum_{n=0}^\infty(-1)^n\left(\frac{1}{4n-a+2}+\frac{1}{4n+a+2}\right)\\
&=&\frac{\pi}{4}\sec\left(\frac{a\pi}{4}\right).
\end{eqnarray}
So
\begin{eqnarray}
I'(a)&=&\int_0^a\frac{\pi}{4}\sec\left(\frac{\pi t}{4}\right)dt\\
&=&\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)-\ln\cos\left(\frac{\pi a}{4}\right)
\end{eqnarray}
and hence
\begin{eqnarray}
I(2)&=&\int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da-\int_0^2\ln\cos\left(\frac{\pi a}{4}\right)da.
\end{eqnarray}
Note
$$ \int_0^2\ln\cos\left(\frac{\pi a}{4}\right)da=-2\ln2 $$
from Evaluating $\int_0^{\large\frac{\pi}{4}} \log\left( \cos x\right) \, \mathrm{d}x $
and it should not be hard to get
$$ \int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=\frac{8G}{\pi}-2\ln 2$$
and thus
$$ I(2)=\frac{8G}{\pi}. $$
$\bf{Update}$ 1: Let us first work on
$ \sum_{n=0}^\infty(-1)^n\left(\frac{1}{4n-a+2}+\frac{1}{4n+a+2}\right)=\frac{\pi}{4}\sec\left(\frac{a\pi}{4}\right)$.
In fact
\begin{eqnarray*}
&&\sum_{n=0}^\infty(-1)^n\left(\frac{1}{4n-a+2}+\frac{1}{4n+a+2}\right)\\
&=&\sum_{n=0}^\infty\left(\frac{1}{8n-a+2}-\frac{1}{8n-a+6}\right)+\sum_{n=0}^\infty\left(\frac{1}{8n+a+2}-\frac{1}{8n+a+6}\right)\\
&=&\sum_{n=0}^\infty\frac{4}{(8n-a+2)(8n-a+6)}+\sum_{n=0}^\infty\frac{4}{(8n+a+2)(8n+a+6)}\\
&=&\sum_{n=-\infty}^\infty\frac{4}{(8n-a+2)(8n-a+6)}=\sum_{n=-\infty}^\infty\frac{4}{(8n-a+4)^2-2^2}\\
&=&\frac{1}{16}\sum_{n=-\infty}^\infty\frac{1}{(n+\frac{4-a}{8})^2-(\frac{1}{4})^2}.
\end{eqnarray*}
Now using a result from Closed form for $\sum_{n=-\infty}^\infty \frac{1}{(z+n)^2+a^2}$ for $a=\frac{i}{4}$ and $z=\frac{4-a}{8}$ and after some basic calculation we can get this result. Also see An alternative proof for sum of alternating series evaluates to $\frac{\pi}{4}\sec\left(\frac{a\pi}{4}\right)$ for a short proof.
$\bf{Update}$ 2: We work on $\int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=\frac{8G}{\pi}-2\ln 2$. In fact
\begin{eqnarray}
\int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=\frac{4}{\pi}\int_0^{\pi/2}\ln\left(1+\sin(a)\right)da=\frac{4}{\pi}\int_0^{\pi/2}\ln(1+\cos(a))da.
\end{eqnarray}
Using $2\cos^2\frac{a}{2}=1+\cos a$ and a result $G=\int_0^{\pi/4}\ln(\cos(t))dt$ from http://en.wikipedia.org/wiki/Catalan%27s_constant, it is easy to obtain
$$ \int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=
\frac{8G}{\pi}-2\ln 2. $$
Plot of the integrand for $a=8$.
As the integrand is even, $I(a) = 2 \int_{0}^\infty \frac{\tanh{(y^2-a^2)}}{y^2-a^2} dy$
We can break the limits of integration into three intervals, $0 < y< a(1-\delta)$, $a(1-\delta) < y< a(1+\delta)$, and $a(1+\delta)<y<\infty$, where $\delta$ is chosen such that $\frac{1}{a} \ll \delta \ll 1$, so that $\tanh{(a \delta)} \approx 1$.
Then, the first part is, $I_1(a) = 2 \int_{0}^{a-a\delta} \frac{\tanh{(y^2-a^2)}}{y^2-a^2} dy \approx 2 \int_{0}^{a-a\delta} \frac{(-1)}{y^2-a^2} dy = \frac{1}{a} \log\frac{2-\delta}{\delta} \approx \frac{1}{a} \log\frac{2}{\delta}$.
The second part is, $I_2(a) = 2 \int_{a-a\delta}^{a+a\delta} \frac{\tanh{(y^2-a^2)}}{y^2-a^2} dy \approx 2 \int_{a-a\delta}^{a+a\delta} \frac{\tanh{(2a(y-a))}}{2a(y-a)} dy = \frac{1}{a} \int_{-2a^2 \delta}^{2a^2 \delta} \frac{\tanh{z}}{z} dz$.
Here I use the known result $\int_0^b \frac{\tanh x}{x} \approx C + \log b$, where $C = \gamma + \log(\frac{4}{\pi})$. Here $\gamma$ is the Euler's constant.
Then, $I_2(a) \approx \frac{2}{a} (C + \log(2 a^2 \delta))$.
The third part is, $I_3(a) = 2 \int_{a+a\delta}^{\infty} \frac{\tanh{(y^2-a^2)}}{y^2-a^2} dy \approx 2 \int_{a+a\delta}^{\infty} \frac{1}{y^2-a^2} dy = \frac{1}{a} \log\frac{2+\delta}{\delta} \approx \frac{1}{a} \log\frac{2}{\delta}$.
After adding everything, $\delta$ cancels, and we get,
$$I(a) = I_1 (a) + I_2(a) + I_3 (a) \approx \frac{2}{a} (C + 2 \log 2 + 2 \log a) = \frac{2\gamma + 2 \log(16/\pi) + 4 \log a}{a}.$$
The numerical value of $\boxed{c_1 = 2 \gamma + 2\log(16/\pi)}$ is approximately $4.41015$.
Best Answer
Partial solution
The function $\frac{\tanh x}{x}$ is analytical near $x=0$, so we can decompose it into the series of $a^{2k}, k=0, 1,2,...$ $$\frac{\tanh(x^2-a^2)}{x^2-a^2}=\frac{\tanh(x^2)}{x^2}-a^2\left(\frac1{x^2\cosh^2(x^2)}-\frac{\tanh(x^2)}{x^4}\right)+O(a^4)$$ Every term is integrable on $(-\infty;\infty)$; considering the two first terms of the asymptotics $$\int_{-\infty}^\infty \frac{\tanh(x^2-a^2)}{x^2-a^2} dx=\int_{-\infty}^\infty \frac{\tanh(x^2)}{x^2}+a^2\int_{-\infty}^\infty\left(\frac{\tanh(x^2)}{x^2}-\frac1{\cosh^2(x^2)}\right)\frac{dx}{x^2}+O(a^4)$$ $$\overset{x^2=t}{=}\int_0^\infty\frac{\tanh t}{t^{3/2}}dt+a^2\int_0^\infty\left(\frac{\tanh x}{x}-\frac1{\cosh^2x}\right)\frac{dx}{x^{3/2}}+O(a^4)$$ $$=\int_0^\infty\frac{\tanh t}{t^{3/2}}dt+a^2\int_0^\infty\left(\tanh x-x\right)\frac{dx}{x^{5/2}}+a^2\int_0^\infty\left(1-\frac1{\cosh^2x}\right)\frac{dx}{x^{5/2}}+O(a^4)$$ Integrating the third term by part $$I(a)=\int_0^\infty\frac{\tanh t}{t^{3/2}}dt+\frac{a^2}2\int_0^\infty\frac{x-\tanh x}{x^{5/2}}dx+O(a^4)\tag{1}$$ where the first term $$\int_0^\infty\frac{\tanh t}{t^{3/2}}dt=\int_0^\infty\left(1-\frac2{e^{2x}+1}\right)\frac{dt}{t^{3/2}}\overset{IBP}{=}8\int_0^\infty\frac{e^{2t}}{(e^{2t}+1)^2}\frac{dx}{\sqrt x}$$ $$=4\sqrt2\int_0^\infty\frac{e^x}{(e^x+1)^2}\frac{dx}{\sqrt x}$$ For the sake of evaluation of the second integral, we consider $$I_1(\alpha)=\int_0^\infty x^\alpha\frac{e^x}{(e^x+1)^2}dx\overset{IBP}{=}\alpha\int_0^\infty\frac{x^{\alpha-1}}{e^x+1}dx=\alpha\Gamma(\alpha)\eta(\alpha)=\Gamma(1+\alpha)\eta(\alpha)\tag{2}$$ Taking $\alpha=-\,\frac12$ $$\int_0^\infty\frac{\tanh t}{t^{3/2}}dt=4\sqrt{2\pi}\,\eta\big(-\frac12\big)\tag{3}$$ Evaluating the second integral in the same way, we get $$\frac{a^2}2\int_0^\infty\frac{x-\tanh x}{x^{5/2}}dx=\frac{16 a^2}{3\sqrt 2}\int_0^\infty\frac{e^{2t}-e^t}{(e^t+1)^3}\frac{dt}{\sqrt t}$$ We consider instead $$I_2(\alpha)=\frac{16 a^2}{3\sqrt 2}\int_0^\infty\frac{e^{2t}-e^t}{(e^t+1)^3}t^\alpha dt$$ Taking a "suitable" $\alpha$ to implement integration by part and consider the terms separatly, in the end we will use the analytical continuation to define the result at $\alpha =-\,\frac12$
Straightforward evaluation gives $$I_2(\alpha)=\frac{16 a^2}{3\sqrt 2}\,\alpha\int_0^\infty t^{\alpha-1}\frac{e^t}{(e^t+1)^2}dt=\frac{16 a^2}{3\sqrt 2}\Gamma(1+\alpha)\eta(\alpha-1)$$ Taking $\alpha =-\,\frac12$ $$\frac{a^2}2\int_0^\infty\frac{x-\tanh x}{x^{5/2}}dx=\frac{16 \sqrt{\pi} a^2}{3\sqrt 2}\eta\big(-\frac32\big)\tag{4}$$ Putting (3) and (4) into (1) and using $\eta(\alpha)=\zeta(\alpha)(1-2^{1-\alpha})$ $$\boxed{\,\,I(a)=-4\sqrt{2\pi}(2\sqrt2-1)\zeta\big(-\frac12\big)-\frac{16\sqrt\pi}3\,\frac{4\sqrt2-1}{\sqrt2}\zeta\big(-\frac32\big)\,a^2+O(a^4)\,\,}$$ Numeric check at WolframAlpha confirms the result.
In the similar way we can evaluate higher terms. Taking the pattern, I would suppose that the full asymptotics could also be obtained, though I don't know the way to get it.