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Problem: Evaluate the integral $$\int{\frac{x^3dx}{1+x^5}}$$

Source: Given to me as a challenge. I can't seem to find a valid substitution for the integral. Also the way should be quick enough to apply in a timed test. I don't know the answer and I have almost given up. I called the challenger and he said that I can ask for help but not look up on Wolfram or any such website as it won't be a proper solution. I don't think it is solvable though :p

My try: Well, I have tried to get around by using many substitutions which make the thing more complicated than it already is. Also I'm looking for the QUICKEST way to approach this problem in case it happens to meet me on a test.

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  • $\begingroup$ Are you able to integrate functions of the form $\frac{1}{\alpha-x}$? Well, then apply a partial fraction decomposition. The integrand function is a meromorphic function with simple poles at $-1$ and at the primitive tenth roots of unity. $\endgroup$ Commented Sep 18, 2017 at 17:22
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    $\begingroup$ Every rational function is integrable. $\endgroup$ Commented Sep 18, 2017 at 17:29
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    $\begingroup$ $$\int \frac{x^3}{1+x^5}dx=\int\frac{\frac{1}{x^2}}{\frac{1}{x^5}+1}dx$$ Now let $\frac 1x=t$. To covert it to $$-\int \frac{1}{t^5+1}$$This is the first thought came to me, I too don't know how to proceed further. $\endgroup$ Commented Sep 18, 2017 at 17:39
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    $\begingroup$ Nah I just phoned the guy and said I'm out...I already looked up the answer now and I was happy to see it...That guy just wasted 2 days of my life for this inhumane problem! $\endgroup$ Commented Sep 18, 2017 at 17:49
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    $\begingroup$ @YourAverageEuler lol it dosent deserve even an hour of yout life $\endgroup$ Commented Sep 18, 2017 at 17:52

2 Answers 2

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Hint: $1+x^5$ factors as $$(1+x)(1-x+x^2-x^3+x^4) = (1+x)(1 + ax + x^2)(1 + bx + x^2)$$ for appropriate $a, b$. Then use partial fractions.

And this is NOT an appropriate question for a timed test.

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  • $\begingroup$ With $a=-\phi$ and $b=\phi-1=\dfrac{1}{\phi}$ where $\phi$ is the golden ratio. Definitely not appropriate for a timed test. One can eliminate the $x^3$ in the numerator by the substitution $x=u^{-1}$. $\endgroup$ Commented Sep 18, 2017 at 18:53
  • $\begingroup$ Will use the approach when I'll have time. Thanks! And yeah it's tedious and boring actually, just mind bending calculations $\endgroup$ Commented Sep 19, 2017 at 7:15
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Note that \begin{align} K(a)&= \int \frac{1-a t}{t^2-2a t+1}dx\\ &=-\frac a2 \ln(t^2-2a t+1)+\sqrt{1-a^2}\tan^{-1}\frac{t-a}{\sqrt{1-a^2}} \end{align} $$\frac5{t^5+1}=\frac1{t+1}+\frac{2(1-t\cos\frac\pi5)}{t^2-2 t\cos\frac\pi5+1}+ \frac{2(1-t\cos\frac{3\pi}5)}{t^2-2t \cos\frac{3\pi}5 +1} $$ and \begin{align} \int \frac1{t^5+1}dt&= \frac15\ln(t+1) +\frac25K(\cos\frac \pi5) +\frac25K(\cos\frac {3\pi}5 ) \end{align} Then, with $x=\frac1t$ $$\int \frac{x^3}{x^5+1}dx= -\int \frac1{t^5+1}dt $$

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