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

Question

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

I'm asked to find the length of the curve $\ln \left(x\right)$ on a certain interval and I get the above integral for my calculation (I have taken out the bounds). So which substitution should work here?

Answer 1

Hyperbolic substitution:

$$\frac1x=\sinh u\implies -\frac1{x^2}dx=\cosh u\;du\implies dx=-\frac{\cosh u\;du}{\sinh^2u}\implies$$

$$\int\sqrt{1+\frac1{x^2}}dx=-\int\sqrt{1+\sinh^2u}\frac{\cosh u}{\sinh^2u}du=\int \tanh^2u\;du=$$

$$=u-\tanh u+K$$

Answer 2

Let $x = \tan\theta$ $$1+\frac{1}{x^2} = \frac{\sec^2\theta}{\tan^2\theta} = \frac{1}{\sin^2\theta}$$ $$\therefore \sqrt{1+\frac{1}{x^2}} = \frac{1}{\sin\theta}$$ Using this substitution, the integral should become much simpler. $$dx = \sec^2\theta d \theta$$

Answer 3

Hint:

Write as

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

Substitute $x=\tan t$ and $dx=\sec^2 t dt.$ Then $\sqrt{x^2+1}=\sqrt{\tan^2 t+1}=\sec t $ and $t=\arctan x$

$$=\int\csc t \sec^2 t dt$$

Using $\color{green}{\sec^2t=\tan^2t+1}$

$$=\int(\tan^2t+1)\csc t dt=\int \tan t\sec t dt+\int \csc t dt$$

Write as:

$$=\int\frac{\sin t}{\cos^2 t}dt+\int csc t dt$$

Substitute $p=\cos t$ and $dp=-\sin t dt$

$$=-\int\frac{1}{p^2}dp+\int \csc t dt$$

And now it is easy to finish.

Answer 4

HINT:

$\sqrt{1+\dfrac1{x^2}}=\dfrac{\sqrt{x^2+1}}{|x|}=$sgn$(x)\cdot\dfrac{x\sqrt{x^2+1}}{x^2}$

Set $\sqrt{x^2+1}=u\implies x^2+1=u^2\implies x\ dx=u\ du$ and $x^2=u^2-1$

Now set $u^2-1=v$

Can you take it from here?