First simplify the logarithmic part.
$$\ln(x^2+1)-2\ln x=\ln\left(\frac{x^2+1}{x^2}\right)$$
$$=\ln\left(1+\frac1{x^2}\right)$$
Therefore,
$$I=\int \frac{\sqrt{x^2+1}}{x^4}\ln\left(1+\frac1{x^2}\right)\,dx$$
Notice that
$$\frac{\sqrt{x^2+1}}{x^4}=\frac1{x}\left(1+\frac1{x^2}\right)^{1/2}\frac1{x^2}$$
Let $$t=\sqrt{1+\frac1{x^2}}$$
Then $$t^2=1+\frac1{x^2}$$
Differentiating,
$$2t\,dt=-\frac{2}{x^3}\,dx$$
$$dt=-\frac1{x^3t}\,dx$$
$$dx=-x^3t\,dt$$
Also, $$\ln\left(1+\frac1{x^2}\right)=\ln(t^2)=2\ln t$$
Substituting in the integral,
$$I=\int \frac{t}{x}\cdot\frac1{x^2}\cdot(2\ln t)\cdot(-x^3t)\,dt$$
$$=-2\int t^2\ln t\,dt$$
Now integrate by parts.
Let $$u=\ln t,\qquad dv=t^2dt$$
Then $$du=\frac{dt}{t},\qquad v=\frac{t^3}{3}$$
Therefore, $$\int t^2\ln t\,dt=\frac{t^3}{3}\ln t-\frac13\int t^2dt$$
$$=\frac{t^3}{3}\ln t-\frac{t^3}{9}$$
Hence,
$$I=-2\left(\frac{t^3}{3}\ln t-\frac{t^3}{9}\right)+C$$
$$=-\frac23t^3\ln t+\frac29t^3+C$$
Factor out
$$\frac13t^3$$
$$I=\frac13t^3\left(\frac23-2\ln t\right)+C$$
Using
$$2\ln t=\ln(t^2)$$
and
$$t^2=1+\frac1{x^2}$$
we get
$$I=\frac13\left(1+\frac1{x^2}\right)^{3/2}\left[\frac23-\ln\left(1+\frac1{x^2}\right)\right]+C$$
Final Answer :
$$\frac13\left(1+\frac1{x^2}\right)^{3/2}\left[\frac23-\ln\left(1+\frac1{x^2}\right)\right]+C$$
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