1. ## Integral

I was doing some practice excersises for some diff eqs and this integral came up...now I got it right, but it was messy, can someone here give a better solution

$\int\sqrt{\frac{1}{x^2}+\frac{1}{x^4}}dx$

Now since the domain is every x greater than zero I rewrote this as

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

Then said, Let $x=\tan(\theta)$

which is not fun, not hard, but not fun.

Is there a cool not obvious (or obvious) trick for this integral?

2. Hi
Originally Posted by Mathstud28
I was doing some practice excersises for some diff eqs and this integral came up...now I got it right, but it was messy, can someone here give a better solution

$\int\sqrt{\frac{1}{x^2}+\frac{1}{x^4}}dx$

Now since the domain is every x greater than zero I rewrote this as

$\int\frac{\sqrt{1+x^2}}{x^2}dx$
Let $x=\sinh t \Longrightarrow \mathrm{d}x=\cosh t \,\mathrm{d}t$

$
\int\frac{\sqrt{1+x^2}}{x^2}\,\mathrm{d}x=\int\fra c{\sqrt{1+\sinh^2t}}{\sinh^2t}\cosh t\,\mathrm{d}t=\int\frac{\cosh^2 t}{\sinh^2 t}\,\mathrm{d}t=\int\frac{1}{\sinh^2 t}+1\,\mathrm{d}t$

hence

$\int\frac{\sqrt{1+x^2}}{x^2}\,\mathrm{d}x=-\frac{1}{\tanh t}+t+C=-\frac{\sqrt{1+x^2}}{x}+\mathrm{asinh} x+C'$

I don't know if it's funnier than your method.

3. Try a reciprocal substitution.

4. Originally Posted by Krizalid
Try a reciprocal substitution.
I am assuming you mean start from here

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

Because that is what most look like when you suggest this method, alright well I will give it a go

Let $x=\frac{1}{u}$

So $dx=\frac{-1}{u^2}$

So we have

$-\int\frac{\frac{1}{u^2}\sqrt{1+\frac{1}{u^2}}}{\fr ac{1}{u^2}}=-\int\sqrt{1+\frac{1}{u^2}}du$

Is this the right start?

5. Yes, now do the remaining algebra and you'll get an easy integral.