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I would like to expand my pool of integral solving skills and thus try to solve older problems again, however with a different method I had used back then, when I encountered them first. For this problem I omitted the lower and upper bound and just compute the indefinite integral (values are not important to me)

One of them was:

\begin{align} & \int \frac{x^2-x+2}{x^3-x^2+x-1}dx = \int \frac{x(x-1)+2}{x^2(x-1)+(x-1)}dx \\[10pt] = {} & \int \frac{x(x-1)}{x^2(x-1)+(x-1)} dx + \int \frac{2}{x^2(x-1)+(x-1)}dx \\[10pt] = {} & \int \frac{x}{x^2+1}dx + \int \frac{2}{(x^2+1)(x-1)} dx \\[10pt] = {} & \frac{1}{2} \log(1+x^2) + \int \frac{2(x+1)}{(x^2+1)(x^2-1)} dx \\[10pt] = {} & \frac{1}{2} \log(1+x^2) + 2 \int \frac{1}{x^4-1} dx + 2 \int \frac{x}{(x^2+1)(x^2-1)} dx \\[10pt] = {} & \frac{1}{4} (\log(1-x^2)-\log(1+x^2)) + \frac{1}{2} \log(1+x^2) + 2 \int \frac{1}{x^4-1} dx \end{align}

However now I am stuck. I am sure, if I looked up in a formula table I would find something to the last integral, but not being able to solve this, means that I am missing some kind of insight or tool to further proceed. Does someone have an idea what to do here, for the last integral?

Any constructive comment, answer is appreciated. As always thanks in advance.

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    $\begingroup$ Why did you explicitly use the "align" environment in MathJax if you weren't going to do any aligning? (I've edited it since then.) $\qquad$ $\endgroup$ Commented Jun 13, 2016 at 16:34
  • $\begingroup$ I mainly wanted to try out putting everything inside one block. Until now I always used the dollar symbols. They seemed a bit inefficient. $\endgroup$ Commented Jun 13, 2016 at 16:36

1 Answer 1

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Be smart, add zeros!

$$I=\int dx\frac{1}{x^4-1}=\\ \int dx\frac{1+\overbrace{x^2-x^2}^{=0}}{(x^2-1)(x^2+1)}=\\\int dx\frac{1}{x^2-1}-\int dx\frac{x^2}{(x^2-1)(x^2+1)}=\\ \int dx\frac{1}{x^2-1}-\int dx\frac{x^2\overbrace{-1+1}^{=0}}{(x^2-1)(x^2+1)}=\\ \int dx\frac{1}{x^2-1}-\int dx\frac{1}{x^2+1}-\underbrace{\int dx\frac{1}{(x^2-1)(x^2+1)}}_{I} $$

therefor

$$ 2I=\int dx\frac{1}{x^2-1}-\int dx\frac{1}{x^2+1}=\text{arctanh}(x)-\arctan(x) $$

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  • $\begingroup$ Nice :P, I really forgot this useful/handy trick/tool. Thanks a lot. This really saves a lot of work. $\endgroup$ Commented Jun 13, 2016 at 16:47
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    $\begingroup$ @Imago u are welcome! $\endgroup$ Commented Jun 13, 2016 at 16:48

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