Integrating sqrt(x^2+1)/x^2 using trig

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SUMMARY

The integral \(\int\frac{\sqrt{x^{2}+1}}{x^{2}} dx\) can be effectively solved using trigonometric substitution. The substitution \(x = \tan(u)\) leads to the integral \(\int (csc(u))^{2} \cdot sec(u) \, du\). Simplifying further by recognizing that \(csc^2(u) = 1 + cot^2(u)\) allows for the integral to be expressed as \(\int \left(sec(u) + cot(u)csc(u)\right) du\). An alternative substitution using hyperbolic functions, specifically \(x = sinh(v)\), can also yield the integral \(\int coth^2(v) \, dv\).

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Homework Statement



[tex]\int[/tex][tex]\frac{\sqrt{x^{2}+1}}{x^{2}} dx[/tex]

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The Attempt at a Solution


I tried a Trig Substitution with x=tan(u) and ended up with [tex]\int (csc(u))^{2} * sec(u) du[/tex]

From here I am kind of stuck. I tried a few different integration by parts methods, but they got very messy. I also couldn't find any sort of table for this.
 
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dtl42 said:
I tried a Trig Substitution with x=tan(u) and ended up with [tex]\int (csc(u))^{2} * sec(u) du[/tex]

Hi dtl42! :smile:

Yes, you're almost there …

Put cosec^2 = 1 + cot^2, and you have:
[tex]\int \left(sec(u)\,+\,cot(u)csc(u)\right) du\,.[/tex] :smile:

(though personally, I'd have used x = sinhv, giving ∫coth^2(v)dv)
 
Thanks very much, I can't believe I didn't see that. Its those variations of the Pythag. Identity that get me.
 

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