A indefinite integral question

Question

The integral is that $$ \int\sqrt{ln\frac{(x+\sqrt{1+x^2})}{1+x^2}}\mathrm{d}x $$I have two questions. The first is that how can I think of that trick:$$\frac{1}{\sqrt{1+x^2}}\mathrm{d}x = \mathrm{d}\left(ln(x+\sqrt{1+x^2})\right) $$Using this trick we can change the original integral into that $$ \int\sqrt{ln(x+\sqrt{1+x^2})}\mathrm{d}\left(ln(x+\sqrt{1+x^2})\right) $$ The following is the next question: I use trigonometric substitution $\tan{u} = x$, and we can change the original integral into that$$\int \frac{\sqrt{ln(\text{tan}u+\text{sec}u})}{\text{sec}u}\mathrm{d}(\text{tan}u) = \int \text{sec}u\sqrt{ln(\text{tan}u+\text{sec}u)} \,\mathrm{d}u$$I find that $\int \text{sec}u \mathrm{d}u = ln(\text{tan}u+\text{sec}u)+C$, can I change ahead integral into that$$\int \text{sec}u\sqrt{\int \text{sec}u\,\mathrm{d}u}\,\mathrm{d}u$$ if this operation is permitted, how can I deal with this integral