We got some very good response to our last week’s problem from several of our “regular” problem-solvers as well as several others who are “new”. There were solutions that were more “algebraic” than others, some that had a more “trigonometric” flavor to them and some that had a combination of both. All the solutions we received this time were correct and they all deserve to be published, but for the sake of brevity I will post just one.

Solution to POW-5: (due to Animesh Datta, Univ of New Mexico)

Note that the given integral may be written as

$\displaystyle \int \frac{x^2 - 1}{x(x^2 + 1) \sqrt{x^2 + 1/x^2}} \, dx$

$\displaystyle = \int \frac{1 - 1/x^2}{(x + 1/x) \sqrt{(x + 1/x)^2 - 2}} \, dx$.

Now, we use the substitution $t = x + 1/x$, which transforms the integral into

$\displaystyle \int \frac1{t \sqrt{t^2 - 2}} \, dt$.

Finally, we use one last (trigonometric) substitution $t = \sqrt{2} \sec \theta$, which transforms the integral into $\displaystyle \int \frac1{\sqrt{2}} \, d\theta$, which evaluates to $\theta /\sqrt{2} + C$, which equals $\displaystyle \frac1{\sqrt2} \arctan \sqrt{\frac12 (x^2 + \frac1{x^2})} + C$. And this is our final answer!

Watch out for the next POW that will be posted by Todd!

Source: I had mentioned earlier that Carl Lira had brought this integral to our attention, and he in turn had found it in the MIT Integration Bee archives. This one was from the year 1994.

Trivia: Four out of the six people who sent correct solutions are either Indians or of Indian origin! Coincidence? 🙂