I tried factoring the denominator and performing partial fraction decomposition, but the algebra becomes too cumbersome for me...

Is there a cleaner approach for this integral?

  • 1
    $\begingroup$ See what Wolfy says wolframalpha.com/input/?i=int%28x%5E2%2F%281%2Bx%5E5%29%2Cx%29 $\endgroup$ Commented Dec 20, 2019 at 19:42
  • $\begingroup$ Based on the comment above, it would seem there is not a cleaner approach... $\endgroup$
    – QC_QAOA
    Commented Dec 20, 2019 at 19:55
  • $\begingroup$ You can use the hypergeometric function, but I guess this isn't what you want. So you probably have to go through the partial fraction route. $\endgroup$
    – bjorn93
    Commented Dec 20, 2019 at 20:11
  • $\begingroup$ From which book is this? Just curious $\endgroup$
    – imranfat
    Commented Dec 20, 2019 at 20:54
  • $\begingroup$ @imranfat It's not an exercise actually. A practice book I've been doing lists it (mistakenly I assume) just after $e^{x^2}$ in its examples of integrals that have no elementary antiderivative. $\endgroup$
    – Riz222
    Commented Dec 21, 2019 at 4:28

2 Answers 2


There is a clean way to integrate. Note that $x^5+1$ factorizes as

$$x^5+1= (1+x)(x^2-2\phi_+x+1)(x^2-2\phi_-x+1)$$

with $\phi_{\pm} = \frac{1\pm\sqrt5}{4}$. So, decompose the integrand,

$$\frac{5x^2}{1+x^5}=\frac1{x+1}-\phi_- \frac{2x+2}{x^2-2\phi_+x+1} -\phi_+ \frac{2x+2}{x^2-2\phi_-x+1}$$

The integral for the first term is just $\ln(x+1)$. The second and third terms have the same form, whose integral can be obtained as,

$$I(x,\phi)= \int \frac{2x+2}{x^2-2\phi x+1}dx =\int \frac{d[(x-\phi)^2] + 2(1+\phi)dx}{(x-\phi)^2 +(1-\phi^2)} $$ $$=\ln\left[(x-\phi)^2 +(1-\phi^2)\right] +\frac{2(1+\phi)}{\sqrt{1-\phi^2}} \tan^{-1}\frac{x-\phi}{\sqrt{1-\phi^2}}$$ $$=\ln\left(x^2-2\phi x+1\right) +2\sqrt{\frac{1+\phi}{1-\phi}} \tan^{-1}\frac{x-\phi}{\sqrt{1-\phi^2}} $$

Thus, the original integral is,

$$\int \frac{x^2}{1+x^5}dx=\frac15\left[\ln(x+1)-\phi_-I(x,\phi_+)-\phi_+I(x,\phi_-)\right] + C$$


Yet it is doable, if you remember that if $\zeta$ is a simple complex pole of the rational function $\dfrac{F(x}{G(x)}$, in the corresponding term in the partial fractions decomposition: $$\frac{F(x)}{G(x)}=\frac{A}{x-\zeta}+\text{other terms, }\enspace \text{we have }\enspace A=\frac{F(\zeta)}{G'(\zeta)}.$$

Now, in the present case, the poles are the fifth roots of $-1$, i.e. $\zeta_k=-\mathrm e^{\tfrac{2ik\pi}5}\quad(k=0,1,\dots 4) $, so we have the (complex) terms: $$\frac{A_k}{x-\zeta_k}, \quad\text{where }\enspace A_k=\frac{\zeta_k^2}{5\,\zeta_k^4}=\frac1{5\,\zeta_k^2}.$$ As the non-real roots are pairwise conjugate, we can group the conjugate terms: $$\frac{A_k}{x-\zeta_k}+\frac{\overline{\! A}_k}{x-\bar \zeta_k}=\frac{(\overline{\! A}_k+A_k)x-(\bar\zeta_kA_k+\zeta_k\,\overline{\! A}_k)}{x^2-(\zeta_k+\bar \zeta_k)x+1}=\frac15\,\frac{\bigl(\zeta_k^2+\bar\zeta_k^2\bigr)x-\bigl(\zeta_k^3+\bar\zeta_k^3\bigr)}{x^2-(\zeta_k+\bar\zeta_k)x+1}.$$

Can you continue?


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