Integrate $1/(u^2 - 1)$ 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.
 A: 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 $$\int\frac1{2x}\,dx=\frac12\ln x$$
A: 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!).
A: 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. 
A: 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,
$$
\int\frac{1}{1-u^2}\,du=-\frac{1}{2}\ln\biggl|\frac{2}{1+u}-1\biggr|+C.
$$
A: 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. 
