$\int \frac{\left(\sqrt{x^2+1}\right)\left\{\ln \left(x^2+1\right)-2 \ln x\right\}}{x^4} dx$ is equal to :

$\frac{1}{9} \frac{\left(x^2+1\right) \sqrt{x^2+1}}{x^3}\left[2-3 \ln \left(\frac{x^2+1}{x^2}\right)\right]+c$

Let $I=\int \frac{\left(\sqrt{\frac{x^2+1}{x^2}}\right) \ln \left(\frac{x^2+1}{x^2}\right)}{x^3} dx$ 

Let $\frac{x^2+1}{x^2}=t$

∴  $-\frac{2}{x^3} dx=dt$

$\Rightarrow I=-\frac{1}{2} \int \sqrt{t} \ln t d t$

$=-\frac{1}{2}\left[(\ln t) . \frac{2 t^{3 / 2}}{3}-\frac{2}{3} \int \frac{1}{t} . t^{3 / 2} d t\right]$

$=\frac{1}{9} t^{3 / 2}[2-3 \ln t]+c$

$=\frac{1}{9} \frac{\left(x^2+1\right) \sqrt{x^2+1}}{x^3}\left[2-3 \ln \left(\frac{x^2+1}{x^2}\right)\right]+c$

Hence (4) is the correct answer.