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

Question

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

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.

Answer 1

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 $x = \sinh t$, $dx = \cosh t \,dt$

Answer 2

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.

Answer 3

Let $x=\sinh t$, then $dx=\cosh t \, dt$

\begin{align*} \int \frac{\sqrt{1+x^{2}}}{x} \,dx &= \int \frac{\cosh t}{\sinh t} \cosh t \, dt \\ &= \int \frac{1+\sinh^{2} t}{\sinh t} \, dt \\ &= \int (\sinh t+\operatorname{csch} t) \, dt \\ &= \cosh t+\int \frac{dt}{2\sinh \frac{t}{2} \cosh \frac{t}{2}} \\ &= \cosh t+\int \frac{\operatorname{sech}^{2} \frac{t}{2} \, dt} {2\tanh \frac{t}{2}} \\ &= \cosh t+\int \frac{d(\tanh \frac{t}{2})} {\tanh \frac{t}{2}} \\ &= \cosh t+\ln \left| \tanh \frac{t}{2} \right|+C \\ &= \cosh t+\ln \left| \frac{\cosh x-1}{\sinh x} \right|+C \\ &= \sqrt{1+x^{2}}+\ln \left| \frac{\sqrt{1+x^{2}}-1}{x} \right|+C \end{align*}