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.

  • $\begingroup$ Personally, I would set z^2=t and either do the series expansion or interpret the integrand as 1F0() Generalized Hypergeometric: en.wikipedia.org/wiki/… 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: dlmf.nist.gov/16.3#E2 $\endgroup$
    – rrogers
    Sep 5, 2021 at 13:49
  • $\begingroup$ take $z = \sinh t $ with $dz = \cosh t \; dt . \; \; \;$ Or $z = \tan w \; \; $ with $dz = \sec^2 w \; dw $ $\endgroup$
    – Will Jagy
    Sep 5, 2021 at 14:14
  • $\begingroup$ Sigh: my dlmf pointer is to the wrong equation the later ones, E3,E4 are usable. $\endgroup$
    – rrogers
    Sep 5, 2021 at 14:31

2 Answers 2



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


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

  • $\begingroup$ Note that your expression for $z$ in terms of $v$ is wrong. The right-hand side of that equation equals $z^2$, not $z$. $\endgroup$
    – Matt L.
    Nov 27, 2021 at 17:07
  • $\begingroup$ @Matt L. I've edited it with regards to your comment, thank you $\endgroup$
    – wd violet
    Nov 28, 2021 at 1:45

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