How to integrate $\int\sqrt{x^2+1}dx$ $$\int\sqrt{x^2+1}dx$$
I've been attempting this problem for days and I've made very little progress. I understand that I should substitute $x=\tan(x)$ and $dx=\sec^2(x)dx$ but beyond that I am very much lost.

You can also do:
$$\int\sqrt{x^2+1}\,dx = \int\frac{x^2+1}{\sqrt{x^2+1}}\,dx = \int x\cdot \frac{x}{\sqrt{x^2+1}}\,dx + \int\frac{1}{\sqrt{x^2+1}}\,dx = x\sqrt{x^2+1} - \int\sqrt{x^2+1}\,dx\,+ \ln(x+\sqrt{x^2+1}) \implies \int\sqrt{x^2+1}\,dx = \frac{1}{2}\big(x\sqrt{x^2+1}+\ln(x+\sqrt{x^2+1})\big) + C $$
