# Finding $\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$

Finding $$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$$

I suppose I need integration by parts and trigo substitution

Let $u=\frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$

Let $dv = \sqrt{1+(\frac{1}{x^2})^2}$, $\frac{1}{x^2} = \tan{\theta}$. Is my substitution OK?

So $x=\frac{1}{\sqrt{\tan{\theta}}} \Rightarrow dx = -\frac{\sec^2{\theta}}{2\sqrt{\tan{\theta}}}\,d\theta$. But this will be very complicated to integrate later?

Am I supposed to be trying something else?

UPDATE: An attempt

$$\int \frac{1}{x} \sqrt{1+\frac{1}{x^4}} dx$$

Let $u = \frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$

Let $dv = \sqrt{1+(\frac{1}{x^2})^2} dx$

Let $\alpha = \frac{1}{x^2} \Rightarrow d\alpha = -\frac{1}{2x^3}$

$dv=\sqrt{1+\alpha^2} d\alpha$

Let $\alpha = \tan{\theta} \Rightarrow d\alpha = \sec^2{\theta} d\theta$

$dv = \sqrt{1+\tan^2{\theta}} \sec^2{\theta} d\theta = \sec^3{\theta}$. Looks wrong here ?

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try $u=\frac{1}{x^2}\Rightarrow du = \frac{-1}{2x^3}$. This will lead to $-\frac{1}{2}\int u^2\sqrt{1+u^2}\,du$ which can be solved by substituting a hyperbolic function –  example Apr 11 '12 at 11:18
substitute : $u=\frac{1}{x^4}$ –  pedja Apr 11 '12 at 11:28
Sry, made a small mistake in my head while writing that comment. See my solution below. –  example Apr 11 '12 at 11:33

$$I = \int \frac{\sqrt{x^4+1}}{x^3} \mathrm{d}x$$

Substitute $u=\sqrt{x^4+1}, \mathrm{d}u=\frac{2x^3}{\sqrt{x^4+1}}\mathrm{d}x$

$$\int \frac{u^2}{2(u^2-1)^{\frac{3}{2}}} \mathrm{d}u$$

Now substitute $u=\sec \theta$ $\mathrm{du} = \sec \theta \tan \theta \mathrm{d}\theta$

Then $(u^2-1)^{\frac{3}{2}} = \tan^3 \theta$

$$I = \frac{1}{2}\int \csc^2 \theta \sec \theta \mathrm{d}\theta = \frac{1}{2}\int (\cot^2 \theta+1) \sec \theta \mathrm{d}\theta$$

And take it from there

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Usually we want to get rid of the square root. One way to do so is to bring in into the form $\sqrt{1+\cosh^2(u)}$ so lets try this in two steps $$\frac{1}{x^2}=v \Rightarrow dv = \frac{-1}{2x^3}dx \Leftrightarrow dx = -2x^3 \, dv$$ $$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx = \int \frac{-2x^3}{x}\sqrt{1+v^2}\,dv = -2\int \frac{1}{v}\sqrt{1+v^2}\,dv$$

With $v=\cosh(u) \Rightarrow dv = \sinh(u)\,du$ and $1+\cosh^2(u) = \sinh^2(u)$ follows $$... = -2\int\frac{\sinh^2(u)}{\cosh(u)} \,du$$

This will become a bit messy from here on. Seeing how the final result is ... $$-\frac{1}{2} \sqrt{1+\frac{1}{x^4}}+\frac{1}{2}\sinh^{-1}\left(x^2\right)$$

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Is $\cosh$ the same as $\arccos$ or $\cos^{-1}$? Also I think $$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx = -2\int \color{red}{\frac{1}{v}}\sqrt{1+v^2}\,dv$$, is wrong? Should it be $\frac{1}{\sqrt{v}}$ –  Jiew Meng Apr 11 '12 at 11:39
@Jiew it is the hyperbolic cosine. en.wikipedia.org/wiki/Hyperbolic_functions . Here we only use $\frac{d}{dx}\cosh(x)=\sinh(x)$, $\frac{d}{dx}\sinh(x)=\cosh(x)$ and $\sinh^2(x)-\cosh^2(x)=1$ though. –  example Apr 11 '12 at 11:41
@Jiew: added 2 more steps in the equations to show that it indeed is $\int\frac{1}{v}...dv$ –  example Apr 11 '12 at 11:43
@Jiwe: ahhh, but I made a mistake. Should be $\sinh^2(u) / \cosh(u)$. Let me see whether I can correct that. –  example Apr 11 '12 at 11:46
Hmm don't see a nice way to do the last steps. Maybe someone can fill them in? –  example Apr 11 '12 at 11:55

\begin{aligned} \int\frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,\mathrm{d}x&=\int \frac{\sqrt{x^4 +1}}{x^3}\,\mathrm{d}x\\ &=\int\frac{x^4 + 1}{x^3\sqrt{x^4+1}}\,\mathrm{d}x\\ &=\int\frac{x}{\sqrt{\left(x^2\right)^2+1}}\,\mathrm{d}x + \int \frac{1}{x^5}\frac{1}{\sqrt{1 + 1/x^4}}\,\mathrm{d}x\\ &=\frac{1}{2}\ln\Big(x^2 + \sqrt{x^4 + 1}\Big) - \frac{1}{2}\sqrt{1+1/x^4} + C\\ &=\frac{1}{2}\ln\Big(x^2 + \sqrt{x^4 + 1}\Big) - \frac{\sqrt{x^4 + 1}}{2x^2}+C \end{aligned}

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