Integral $\frac{\sqrt{1+x^2}}{x}$

Question

I understand that I will be using trig substitution, and tangent will be what is used, but I get confused later down the road when integrating with the trig.

Answer 1

If you want to use a trig substitution, take $x=\tan\left(u\right) $ to get $dx=\sec^{2}\left(u\right) $, $\sqrt{1+x^{2}}=\sqrt{1+\tan^{2}\left(u\right)}=\sec\left(u\right) $ and $x=\tan^{-1}\left(u\right) $. Then $$\int\frac{\sqrt{1+x^{2}}}{x}dx=\int\csc\left(u\right)\sec^{2}\left(u\right)du=\int\csc\left(u\right)\left(1+\tan^{2}\left(u\right)\right)du= $$ $$=\int\frac{\sin\left(u\right)}{\cos^{2}\left(u\right)}du+\int\csc\left(u\right)du $$ and I think you can get it from here.