I want to evaluate the following indefinite integral with IBP and also with the limits $0$ to $\infty$. Once indefinite integral is calculated, it is easy to evaluate it within given limits.
$$I=\int \frac{x^{4n-2}}{x^{4n}+x^{2n}+1} \, \mathrm dx$$
So, I proceed through Partial fractions method.
$$I=\int \frac{x^{4n-2}}{x^{4n}+x^{2n}+1} \, \mathrm dx=\frac12
\int \frac{x^{2n-2}+x^{n-2}}{x^{2n}+x^n+1} \, \mathrm dx+\frac12
\int \frac{x^{2n-2}-x^{n-2}}{x^{2n}-x^n+1} \, \mathrm dx$$
We can split it into $4$ integrals but I think that would make it a little messy.
Let $\space I=I_{1}+I_{2}$
$$I_1= \int \frac{1}{2x}\bigg(\frac{2x^{2n-1}+x^{n-1}-x^{2n-1}}{x^{2n}+x^{n}+1}\bigg)\mathrm dx$$
$$\implies I_1=\int \frac{1}{2nx}\bigg(\frac{2nx^{2n-1}+nx^{n-1}}{x^{2n}+x^n+1}-\frac{nx^{2n-1}}{x^{2n}+x^n+1}\bigg) \, \mathrm dx$$
After IBP, we get
$$I_1=\frac {1}{2n}\Bigg[\frac{1}{x}\ln(x^{2n}+x^n+1)+\int \frac {1}{x^2} \ln(x^{2n}+x^n+1) \, \mathrm dx\Bigg]-\frac12 \int \frac{x^{2n-2}}{x^{2n}+x^{n}+1} \, \mathrm dx$$
Let $x=\frac{1}{x}$ for above two integrals and we get
$$I_1 = \frac {1}{2nx} \ln(x^{2n}+x^n+1)+x\ln(x)-x-\frac{1}{2n}\int \ln(x^{2n} + x^{n}+1)\,\mathrm dx +\frac12 \int \frac{1}{x^{2n}+x^n+1} \, \mathrm dx$$
Same thing goes with $I_2$ and we are stuck with the
following integrals
$$\int \ln(x^{2n}\pm x^{n}+1)\,\mathrm dx \space and \int \frac{1}{x^{2n}\pm x^n+1} \, \mathrm dx$$
I don't think so that their anti-derivative can be expressed in terms of elementary functions. But I'm also interested in definite integrals of those functions i.e
$$\int_0^\infty \ln(x^{2n}+x^{n}+1)\mathrm dx \text{ and } \int_0^\infty \frac{1}{x^{2n}+x^n+1}\,\mathrm dx$$
Best Answer
It is actually easier to evaluate the definite integral directly. For $m>1$ \begin{align} I(m)=& \int_0^\infty \frac{1}{x^{2m}+x^m+1}dx\\ =& \>\frac1{2i\sin a }\int_0^\infty \overset{t=xe^{i\frac am}}{\frac{e^{i a}}{x^m e^{i a}+1}} -\overset{t=xe^{-i\frac am}}{\frac{e^{-i a}}{x^m e^{-i a}+1} }\>dx\>\>\>\>\>\> (a=\frac\pi3 )\\ =& \>\frac{\sin\frac{(m-1)\pi}{3m}}{\sin\frac\pi3}\int_0^\infty \frac1{t^m+1}dt = \frac{2\pi\sin\frac{(m-1)\pi}{3m}}{\sqrt3 m \sin\frac\pi m} \end{align}
where $\int_0^\infty \frac{dt}{t^m+1}= \frac\pi{m\sin\frac\pi m}$. Then
\begin{align} \int_0^\infty \frac{x^{4n-2}}{x^{4n}+x^{2n}+1} dx \overset{x\to\frac1x} =\> \int_0^\infty \frac{1}{x^{4n}+x^{2n}+1} dx= I(2n)=\frac{\pi \cos\frac{(n+1)\pi}{6n} }{\sqrt3 n \sin\frac\pi {2n}} \end{align}