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Today, the CalcBee sample problems got released. The following problem was my creation and I wanted to see how many solutions people can come up with. The result is very beautiful and I thought it would be instructive to see multiple solutions to this if they exist. Note that, on the real contest, none of the problems will be this hard. This was meant as a challenge since partial fractions do not exactly work right away.

Find $ \displaystyle\int \frac {1-x^2}{1+3x^2+x^4} \, \mathrm{d}x $.

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  • $\begingroup$ Indeed, partial fractions do not seem to get you the answer you are looking for. It gives you this, but if you try to solve it manually you get this and this $\endgroup$
    – user1537366
    Dec 25, 2014 at 9:09

3 Answers 3

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First substitute $x=e^y$ ("$dx=e^y dy$"): $$\int\frac{1-x^2}{1+3x^2+x^4}dx=\int\frac{e^y-e^{3y}}{1+3e^{2y}+e^{4y}}dy=\int\frac{e^{-y}-e^y}{e^{-2y}+3+e^{2y}}dy=\int\frac{-2\sinh y}{(2\cosh y)^2+1}dy$$ Now substitute $z=2\cosh y$ ("$dz = 2\sinh y\,dy$"): $$\int\frac{-2\sinh y}{(2\cosh y)^2+1}dy=\int\frac{-1}{z^2+1}dz=\text{arccot}\,z$$ So the answer should be $$\text{arccot}(2\cosh y)=\text{arccot}(e^y+e^{-y})=\text{arccot}(x+1/x)$$ and it indeed is, as can be verified by taking the derivative.

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HINT:

Divide the numerator & the denominator by $x^2$

and as $\int\left(1/x^2-1\right)dx=-x-\dfrac1x$

set $x+\dfrac1x=u$ and $\dfrac1{x^2}+3+x^2=\left(x+\dfrac1x\right)^2-2+3$

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  • $\begingroup$ @AhaanSRungta, So, the answer will be $$-\arctan\left(x+1/x\right)+C $$ $$=-\arctan\frac{x^2+1}x+C=-\text{arccot}\frac x{1+x^2}+C$$ $$=-\left(\frac\pi2-\arctan\frac x{1+x^2}\right)+C$$ $\endgroup$
    – lab bhattacharjee
    Dec 20, 2014 at 16:36
  • $\begingroup$ Correct! When we update the PDF, do you mind if we credit you with the alternate solution? $\endgroup$
    – Ahaan S. Rungta
    Dec 20, 2014 at 16:38
  • $\begingroup$ @AhaanS.Rungta, Not sure if I've understood you $\endgroup$
    – lab bhattacharjee
    Dec 20, 2014 at 16:39
  • $\begingroup$ When we update the PDF, do you find if we add this alternate solution and credit you? $\endgroup$
    – Ahaan S. Rungta
    Dec 20, 2014 at 16:41
  • $\begingroup$ @AhaanS.Rungta, No problem:) $\endgroup$
    – lab bhattacharjee
    Dec 20, 2014 at 16:42
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For the record, for ease of access, here is my solution.

Notice that the degree of the denominator is double the degree of the numerator, so there is a possibility that the answer is of the form $ \arctan \left( f(x) \right) $, where $f(x)$ is a rational function. So we try it out. If this is the case, we have $$ \frac {f'(x)}{1 + \left[ f(x) \right]^2} = \frac {1-x^2}{1+3x^2+x^4}. $$Then, let $ f(x) = \frac {p(x)}{q(x)} $, where $p(x)$ and $q(x)$ are polynomials with degree $a$ and $b$, respectively. Then, $$ f'(x) = \frac {q(x) p'(x) - p(x) q'(x)}{\left[ q(x) \right]^2}, $$where the numerator has degree less than or equal to $a+b-1$ and the denominator has degree $2b$. Therefore, $b=2$ and $a=1$. Then, $$ f(x) = \frac {rx + s}{tx^2 + ux + v} \implies f'(x) = \frac {(rv-su) - 2stx - rtx^2}{t^2 x^4 + 2tux^3 + (2tv+u^2) x^2 + 2uvx + v^2}. $$Then, the derivative of $ \arctan \left( f(x) \right) $ is $$ \frac {f'(x)}{1 + \left[ f(x) \right]^2} = \frac {\frac {(rv-su) - 2stx - rtx^2}{t^2 x^4 + 2tux^3 + (2tv+u^2) x^2 + 2uvx + v^2}}{1 + \left( \frac {rx+s}{tx^2 + ux + v} \right)^2}. $$We find that $p(x)=x$ and $q(x)=1+x^2$, so our answer is $$ \boxed {\arctan \left( \frac {x}{1+x^2} \right)}. $$

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