First, observe that:
\[
\frac{\sqrt{x^2 + 1}}{x^2} = \frac{1}{x} \cdot \frac{\sqrt{x^2 + 1}}{x}
\]
Thus, we can rewrite the integral as:
\[
J = \int \left( \frac{\sqrt{x^2 + 1}}{x} \log(x^2 + 1) - 2 \cdot \frac{\sqrt{x^2 + 1}}{x} \log x \right) \, dx
\]
Now, break it into two parts:
\[
J_1 = \int \frac{\sqrt{x^2 + 1}}{x} \log(x^2 + 1) \, dx
\]
\[
J_2 = \int -2 \cdot \frac{\sqrt{x^2 + 1}}{x} \log x \, dx
\]
Each of these integrals can be solved using integration by parts or substitution. The final result for both integrals can be obtained as:
\[
J = J_1 + J_2
\]