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So I have this integral

$$\int \frac{1}{\sqrt{(z^2+1)^3}}dz$$

I tried substituting $z^2+1=t$

but I just get a more complicated integral. WolframAlpha solved the integral really quickly by substituting $z=\tan{u}$, by which the integral transforms into the integral of cosine.

My question is, is there any other way to solve this integral? This was pretty easy, but this substitution would have never occurred to me. This is the first time I am seeing this kind of substitution being used for non-trigonometric integrals.

l0ner9
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  • Personally, I would set z^2=t and either do the series expansion or interpret the integrand as 1F0() Generalized Hypergeometric: https://en.wikipedia.org/wiki/Generalized_hypergeometric_function#The_series_1F0 and then integrate the terms; the two forms are the same except that there are more "rules" for the Generalized Hypergeometric form, and it returns hypergeometric forms which are similar to what you started with: https://dlmf.nist.gov/16.3#E2 – rrogers Sep 05 '21 at 13:49
  • take $z = \sinh t $ with $dz = \cosh t ; dt . ; ; ;$ Or $z = \tan w ; ; $ with $dz = \sec^2 w ; dw $ – Will Jagy Sep 05 '21 at 14:14
  • Sigh: my dlmf pointer is to the wrong equation the later ones, E3,E4 are usable. – rrogers Sep 05 '21 at 14:31

2 Answers2

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Note

$$\int \frac{1}{\sqrt{(z^2+1)^3}}dz = \int \frac{\frac1{z^3}}{\sqrt{(1+\frac1{z^2})^3}}dz = \int \frac{-\frac12d(1+\frac1{z^2})}{\sqrt{(1+\frac1{z^2})^3}} =\frac1{\sqrt{1+\frac1{z^2}}}+C $$

Quanto
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$$\int \frac{1}{\sqrt{(z^2+1)^3}}dz $$ Let $ v=\displaystyle\frac{1}{\sqrt{z^2+1}} $ so $ \displaystyle dv={\frac{-zdz}{(z^2+1)^{\frac{3}{2}}}} $

The integral becomes $$ \int\frac{-dv}{z} $$ But $z=\displaystyle \frac{\sqrt{1-v^2}}{v}$

$$ \int{\frac{-v}{\sqrt{1-v^2}}dv} $$ Letting $u=1-v^2$ gives $$ \frac{1}{2}\int{\frac{1}{\sqrt{u}}}du $$ $$ \Longrightarrow\sqrt{1-v^2} $$ From here, return $v$

wd violet
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