Evaluating $\int\sqrt{1 + \frac1{x^2}}\mathrm dx$

Question

How do I solve this integral? $$\int \sqrt{1 + \frac1{x^2}}\mathrm dx$$

I tried substituting $x=\cot t$ but ended up getting another tricky integral, so any help would be appreciated.

Here's a solution via Wolfram|Alpha with all steps explained--but perhaps an answer by a user here will be more informative. It doesn't look very pretty, regardless. — Daniel W. Farlow, Feb 14 '15 at 08:46

score 0 · Answer 1 · answered Feb 14 '15 at 08:44

0

Hint: Use $x = \tan \theta $. Can you take it from here?

answered Feb 14 '15 at 08:44

DeepSea

77,651

score 0 · Answer 2 · answered Feb 14 '15 at 09:03

0

Hint :$$I=\int \frac{x}{\sqrt{x^2+1}}dx+\int \frac{1}{x\sqrt{x^2+1}}dx,\\ \int \frac{1}{x\sqrt{x^2+1}}dx=-\int\frac{d(\frac{1}{x})}{\sqrt{1+\frac{1}{x^2}}}$$

answered Feb 14 '15 at 09:03

Samrat Mukhopadhyay

16,277

Evaluating $\int\sqrt{1 + \frac1{x^2}}\mathrm dx$

2 Answers2