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I'm stuck on this question of evaluating an integral of the type:

$$\int \dfrac{x^{2n+1}}{x^{2k+1}+1} \,dx$$

with $k > n$. More concretely I try to compute:

$$\int_1^2 \dfrac{x^{3}}{x^{5}+1} \,dx.$$

Firstly, I split $x^5 + 1$ into parts and tried doing it and it didn't work.

Second, I tried converting the numerator to the form of $a(x^5+1)$ but it didn't work.

Lastly, I checked it using a calculator (online) but it showed a long lengthy answer which I couldn't understand. Please help me.

Thanking you in advance,

Arnav

Michael Hardy
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ANN_Techy
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    It is a huge expression with values of $\arctan$'s and $\log$'s. Whether you can understand it or not, it is what it is. – Gary Mar 03 '22 at 12:14
  • In general, $\int \frac{x^n}{x^k+1} , dx = \frac{x^{n+1}}{n+1} {}_2 F_1 \left(1,\frac{n+1}{k};1+\frac{n+1}{k}; -x^k\right)+C$. It’s unrealistic to expect manageable answers in general. – KStarGamer Mar 03 '22 at 12:21
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    An alternative to @KStarGamer's notation: substituting $x=\tan^{2/(2k+1)}t,,y=\sin^2t$ and using $\sin^2\arctan z=\frac{z^2}{1+z^2}$,$$\begin{align}\int_a^b\frac{x^{2n+1}dx}{x^{2k+1}+1}&=\frac{1}{2k+1}\int_{a^{2k+1}/(1+a^{2k+1})}^{b^{2k+1}/(1+b^{2k+1})}y^{(2n+2)/(2k+1)-1}(1-y)^{-(2n+2)/(2k+1)}dy\&=\frac{1}{2k+1}\left[B(x;,\tfrac{2n+2}{2k+1},,\tfrac{2k-2n-1}{2k+1})\right]_{a^{2k+1}/(1+a^{2k+1})}^{b^{2k+1}/(1+b^{2k+1})}.\end{align}$$ – J.G. Mar 03 '22 at 12:26
  • Where you say you split $x^5+1$ into parts, do you mean that you factored it as one first-degree polynomial and two irreducible quadratic polynomials? – Michael Hardy Mar 03 '22 at 16:14

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Let $x=\frac{t-1}{1+t}$ and decompose the resulting integrand into manageable terms

\begin{align} I=&\int_1^2 \dfrac{x^{3}}{x^{5}+1} dx = \frac15\int^{\frac13}_0\frac{(1+3t^2)+(t^2+3)t}{t^4+2t^2+\frac15}dt\\ ={}&\frac1{10}\int^{\frac13}_0 \bigg(\frac{3+\sqrt5}{t^2+1+\frac2{\sqrt5}} + \frac{3-\sqrt5}{t^2+1-\frac2{\sqrt5}}\bigg)dt +\frac1{10}\int^{\frac19}_0\frac{y+3}{y^2+2y+\frac15}dy \end{align} where an additional substitution $y=t^2$ is made in the last integral. Then, each of the three integrals above can be readily computed.

Quanto
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