$\int \frac{\sqrt{x^2+1}[\log (x^2+1)-2\log x]}{x^4}dx=$

$I=\int \frac{\sqrt{x^2+1}[\log (x^2+1)-2\log x]}{x^4} dx$

Taking $x^2$ common from $\sqrt{x^2+1}$

$=\int \frac{{\left(x^2\right)}^{\frac{1}{2}}{\left(1+\frac{1}{x^2}\right)}^{\frac{1}{2}}(\log (x^2+1)-\log x^2)}{x^4} dx$ $[n\log m=\log m^n]$

$=\int \frac{x{\left(1+\frac{1}{x^2}\right)}^{\frac{1}{2}}\left(\log \frac{x^2+1}{x^2}\right)}{x^4}dx \left[\log m-\log n=\log \frac{m}{n}\right]$

$=\int \frac{{(1+\frac{1}{x^2})}^{\frac{1}{2}}\left(\log \left(1+\frac{1}{x^2}\right)\right)}{x^3}$

Let $t=1+\frac{1}{x^2}\Rightarrow dt=\frac{-2}{x^3}$

$\Rightarrow -\frac{dt}{2}=\frac{dx}{x^3}$

$I=\int t^{\frac{1}{2}}\log \left(t\right) \left(\frac{-1}{2}dt\right)$

$=\left(\frac{{\displaystyle -1}}{{\displaystyle 2}}\right)\int \left(\log t\right)\sqrt{t}dt$

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

$=\frac{-1}{2}\left[\frac{{\displaystyle 2}}{{\displaystyle 3}}\left(\log t\right){\text{\hspace{0.17em}t}}^{3/2}-\frac{{\displaystyle 2}}{{\displaystyle 3}}\int t^{1/2}dt\right]$

$=\frac{-1}{2}\left[\frac{{\displaystyle 2}}{{\displaystyle 3}}\left(\log t\right).t^{3/2}-\frac{{\displaystyle 4}}{{\displaystyle 9}}{t}^{3/2}\right]+C$

$=\frac{1}{2}\frac{2}{3}{t}^{3/2}\left[\frac{{\displaystyle 2}}{{\displaystyle 3}}-\log t\right]+C$

$=\frac{1}{3}{\left(1+\frac{{\displaystyle 1}}{{\displaystyle x^2}}\right)}^{3/2}\left[\frac{{\displaystyle 2}}{{\displaystyle 3}}-\log \left(1+\frac{{\displaystyle 1}}{{\displaystyle x^2}}\right)\right]+C$

$=\frac{1}{9}{\left(1+\frac{{\displaystyle 1}}{{\displaystyle x^2}}\right)}^{3/2}\left[2-3\log \left(1+\frac{{\displaystyle 1}}{{\displaystyle x^2}}\right)\right]+C$