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I was viewing past questions of the MIT Integration Bee and came across the problem:

$$ \int \frac{x^2+1}{x^4-x^2+1}\,dx$$

According to MIT, first semester calculus should be sufficient enough to solve this problem.

I have tried many conventional methods, unsuccessfully, and would appreciate any and all help.

Edit: Changed LaTeX formatting

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2 Answers 2

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$$F=\frac{x^2+1}{x^4-x^2+1}=\frac{1+\dfrac1{x^2}}{x^2-1+\frac1{x^2}}$$

Observe that $\displaystyle\int\left(1+\dfrac1{x^2}\right)=x-\frac1x$

So,$\displaystyle F=\frac{1+\dfrac1{x^2}}{\left(x-\frac1x\right)^2+2-1}$

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  • $\begingroup$ While the other answer was the one that immediately came to my mind, this one is quite elegant. :) $\endgroup$
    – apnorton
    Jun 7, 2014 at 17:14
  • $\begingroup$ So the answer should then be arctan(x-(1/x)), right? $\endgroup$
    – A is for Ambition
    Jun 7, 2014 at 17:17
  • $\begingroup$ @user155812, Elementary formula, right? $\endgroup$
    – lab bhattacharjee
    Jun 7, 2014 at 17:20
  • $\begingroup$ Yes, it would seem like it. Also, WolframAlpha says the answer is arctan(x/(1-x^2)). Wolfram's answer and your answer are slightly different functions with different domains. In this case, which answer is the "correct" or acceptable answer? $\endgroup$
    – A is for Ambition
    Jun 7, 2014 at 17:22
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    $\begingroup$ @user155812, $$\arctan\left(x-\frac1x\right)=\arctan\frac{x^2-1}x=-\arctan\left(\frac{1-x^2}x\right)$$ $$=\text{arccot}\left(\frac{1-x^2}x\right)-\frac\pi2=\arctan\frac x{1-x^2}-\frac\pi2$$ $\endgroup$
    – lab bhattacharjee
    Jun 8, 2014 at 5:35
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Maybe not as elegant but more intuitive, there's always partial fractions.

$$\int\frac{x^2+1}{x^4-x^2+1}dx=\int\frac{x^2+1}{x^4+2x^2+1-3x^2}dx=\int\frac{x^2+1}{(x^2+1+x\sqrt3)(x^2+1-x\sqrt3)}dx=$$ $$\frac12\int\frac{dx}{(x+\frac{\sqrt3}2)^2+\frac14}+\frac12\int\frac{dx}{(x-\frac{\sqrt3}2)^2+\frac14}=$$ $$\int\frac{2dx}{(2x+\sqrt3)^2+1}+\int\frac{2dx}{(2x-\sqrt3)^2+1}=$$ $$\tan^{-1}(2x+\sqrt3)+\tan^{-1}(2x-\sqrt3)$$

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  • $\begingroup$ Thanks for the feedback. I enjoyed your solution because the answer you achieved would also work better if the original question were a definite integral, rather than an indefinite integral. This is because the other answer has a domain excluding 0, whereas your answer does not. So, for instance, if the question were a definite integral with 0 as one of the limits of integration, only your answer could be used to compute a real value. $\endgroup$
    – A is for Ambition
    Jun 7, 2014 at 20:43

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