# Evaluating the integral $\int \sqrt{1 + \frac{1}{x^2}} dx$

$$\int \sqrt{1 + \frac{1}{x^2}} dx$$

This is from the problem calculating the arc length of $y=\log{x}$.

I tried $x = \sinh{t}$ or $\frac{1}{x} = \tan{t}$ but all failed.

• Actually, $\frac{1}{x}=\tan t$ or, equivalently, $x=\cot t$ will work fine and, as Travis points out, so will $x=\sinh t$. – user84413 Feb 20 '16 at 0:30

Hint (For $x > 0$) we can rewrite this as $$\int \frac{\sqrt{1 + x^2}}{x} dx.$$ The radical expression $\sqrt{1 + x^2}$ suggests using the substitution $x = \tan \theta$, $dx = \sec^2 \theta \,d\theta$ or the substitution $x = \sinh t$, $dx = \cosh t \,dt$
If we wish, we can avoid trigonometric or hyperbolic function substitution. For we want $$\int \frac{x\sqrt{x^2+1}}{x^2}\,dx.$$ Let $u^2=x^2+1$. Then $u\,du=x\,dx$, and our integral becomes $$\int \frac{u^2}{u^2-1}\,du.$$ Note that $\frac{u^2}{u^2-1}=1+\frac{1}{u^2-1}$, and use partial fractions.
Let $x=\sinh t$, then $dx=\cosh t \, dt$