Could someone please show me how to evaluate this integral (maybe doing all the steps)? $\int_0^{\sqrt{3}}{\frac{\sqrt{1+x^2}}{x}}\,dx$ I prefer if you avoid to follow the same method used by WolframAlpha (with $\csc$, $\sec$ ecc).
This is what I tried 'till now:
- Substitution with $\sqrt{1+x^2} = u$ I obtained: $\int{\frac{u}{\sqrt{u^2-1}}\frac{u}{\sqrt{u^2-1}}}\,du = \int{\frac{u^2}{u^2-1}}\,du$ But not knowing how to continue, I tried another substitution with $u^2 - 1 = s$ and I obtained: $\int{\frac{s+1}{s} \frac{1}{2\sqrt{s+1}} }\,ds = \frac{1}{2}\int{\frac{s+1}{s\sqrt{s+1}}}\,ds$ But, again, not knowing how to continue I decided to ask here.
Thanks in advance for the help!