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Is there any way possible that I might integrate $$ \int\frac{1}{u^2-1}\,du $$ without appealing to partial fraction decomposition?

I am trying to work some interesting $u$-substitution integrals with novice students who do not need to be taught partial fraction decomposition.

Thank you.

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    $\begingroup$ Why don't they need to be taught PFD? $\endgroup$ – Andrew Li Apr 14 '18 at 16:07
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    $\begingroup$ consider trigonometry! $\endgroup$ – Karn Watcharasupat Apr 14 '18 at 16:07
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    $\begingroup$ Hint: et $u = \sec t$. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 14 '18 at 16:07
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    $\begingroup$ IMO, substitution is harder than partial fraction decomposition, which is just a matter of algebra, not calculus. $\endgroup$ – Yves Daoust Apr 14 '18 at 16:20
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    $\begingroup$ $u=\cosh x$ probably works just as well as $\sec x$, though I suppose hyperbolic functions may not be taught yet $\endgroup$ – John Doe Apr 14 '18 at 16:25
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The moment you see $$u^2-1$$ you should think of some trigonometric stuff.

So if you remember,

$$\tan^2x=\sec^2x-1$$

thus $$u=\sec x$$ surely is a candidate (and there are a lot of other variations to play around with!).

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  • $\begingroup$ How would you go about integrating $\csc x$? This is what this would lead to. If I was given that, I'd probably end up coming back to this and using partial fractions.. $\endgroup$ – John Doe Apr 14 '18 at 16:10
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    $\begingroup$ @JohnDoe It's just $$-\ln(\csc x + \cot x)+C$$ You can always teach your student how this was derived since it's pretty manageable. $\endgroup$ – Karn Watcharasupat Apr 14 '18 at 16:12
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Use trigonometric substitution as the integrand contains $u^2-a^2$ where $a=1$. Substitute $u = a\sec \theta$, or just $u = \sec \theta$.

$$u = \sec \theta\quad \mathrm du = \sec\theta\tan\theta\,\mathrm d\theta$$ $$\int {1\over \sec^2 \theta - 1} \sec\theta\tan\theta\,\mathrm d\theta$$

Which can be integrated by remembering the identity $\tan^2 \theta + 1 = \sec^2 \theta$:

$$\int {1\over \sec^2 \theta - 1} \sec\theta\tan\theta\,\mathrm d\theta = \int {\sec \theta \over \tan\theta}\mathrm d\theta = \int \csc \theta \,\mathrm d\theta$$

And you can find the antiderivative of $\csc \theta$ via identities, see this. Of course, this would be much easier via partial fraction decomposition.

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This is motivated by the solution, so it works quite nicely. It is an alternative to the $\sec$ substitution that has been suggested.$$u=\frac{1-x}{1+x}$$ $$\frac1{u^2-1}=-\frac{4x}{(1+x)^2}\,\,\,\,,\,\,\,\,\,\,\,\,du=-\frac2{(1+x)^2}\,dx\\$$ so the integral becomes $$\int\frac1{2x}\,dx=\frac12\ln x$$

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We have $$ \frac{1}{1-u^2}=\frac{1}{(1+u)(1-u)}=\frac{1}{(1+u)(2-(1+u))}=\frac{1}{(1+u)^2\bigl(\frac{2}{1+u}-1\bigr)}. $$ Thus, $$ \int\frac{1}{1-u^2}\,du=-\frac{1}{2}\ln\biggl|\frac{2}{1+u}-1\biggr|+C. $$

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The answers provided already show many different ways this integral can be approached without Partial Fractions. Here is one more. Similar to the trigonometric substitution $u = \sec(x)$ we can employ the hyperbolic substitution $u = \tanh(x)$

\begin{align} \int \frac{1}{u^2 - 1}\:du &= \int \frac{1}{\tanh^2(x) - 1}\cdot -\operatorname{sech}^2(x)\:dx \\ &= \int \frac{1}{\operatorname{sech}^2(x)}\cdot -\operatorname{sech}^2(x)\:dx \\ &= -x + C = -\operatorname{arctanh(u)} + C \end{align}

Where $C$ is the constant of integration.

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