Two ways to integrate

Back in high school, we were taught to integrate $\frac{1}{1-x^2}$ by decomposing it into partial fractions, while $\frac{1}{1+x^2}$ was to be integrated using trigonometric substitution ($x = \tan{\theta}$). It bothered me how the two methods were so vastly different just because of a change of sign. But hey, who says we can’t integrate $\frac{1}{1+x^2}$ with partial fractions?

$\begin{aligned} \int \! \frac{1}{1+x^2} \mathrm{d}x &= \int \! \frac{1}{1-(ix)^2} \mathrm{d}x \\ &= \frac{1}{2} \int \! \frac{1}{1-ix} + \frac{1}{1+ix} \mathrm{d}x \\ &= \frac{1}{2} \left[\frac{1}{-i} \ln{(1-ix)} + \frac{1}{i} \ln{(1+ix)}\right] + C\\ &= \frac{1}{2} \left[i\ln{(1-ix)} - i\ln{(1+ix)}\right] + C\\ &= \frac{i}{2} \ln{\frac{1-ix}{1+ix}} + C \end{aligned}$\begin{aligned} \int \! \frac{1}{1+x^2} \mathrm{d}x &= \int \! \frac{1}{1-(ix)^2} \mathrm{d}x \\ &= \frac{1}{2} \int \! \frac{1}{1-ix} + \frac{1}{1+ix} \mathrm{d}x \\ &= \frac{1}{2} \left[\frac{1}{-i} \ln{(1-ix)} + \frac{1}{i} \ln{(1+ix)}\right] + C\\ &= \frac{1}{2} \left[i\ln{(1-ix)} - i\ln{(1+ix)}\right] + C\\ &= \frac{i}{2} \ln{\frac{1-ix}{1+ix}} + C \end{aligned}

This may look like a strange beast, but it really is $\tan^{-1}{x} + C$ How do we show this? Based on Euler’s formula, we can write $\tan{x}$ as $-i\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{ix}}$. Substitute that into the integrated expression and see for yourself!

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Jez Ng 06 January 2011
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