# Integrate $\displaystyle\int\frac{1}{u^2 - 1}\,du$ without partial fractions?

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.

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

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!).

• 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.. – John Doe Apr 14 '18 at 16:10
• @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. – Karn Watcharasupat Apr 14 '18 at 16:12

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.

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 $$\int\frac1{2x}\,dx=\frac12\ln x$$

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, $$\int\frac{1}{1-u^2}\,du=-\frac{1}{2}\ln\biggl|\frac{2}{1+u}-1\biggr|+C.$$

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.