$\int \frac{x^{5m-1}+2x^{4m-1}}{{\left(x^{2m}+x^m+1\right)}^3}dx=f\left(x\right)+c$

$\Rightarrow \int \frac{x^{5m-1}+2x^{4m-1}}{x^{6m}{\left(1+x^{-m}+x^{-2m}\right)}^3}dx=f\left(x\right)+c$

$\Rightarrow \int \frac{x^{-m-1}+2x^{-2m-1}}{{\left(1+x^{-m}+x^{-2m}\right)}^3}dx=f\left(x\right)+c$.

If, $1+x^{-m}+x^{-2m}=t$

$\Rightarrow -m\left(x^{-m-1}+2x^{-2m-1}\right)dx=dt$

$\therefore \int \frac{-dt}{m{t}^3}=f\left(x\right)+c$

$\Rightarrow -\frac{1}{m}\int t^{-3}dt=f\left(x\right)+c$

Using $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C, n\ne -1$

$\Rightarrow -\frac{1}{m}\left(\frac{t^{-2}}{-2}\right)=f\left(x\right)+c$

$\Rightarrow \frac{1}{2m{\left(1+x^{-m}+x^{-2m}\right)}^2}+c=f\left(x\right)+c$

$\Rightarrow \frac{1}{2m{\left(1+{\displaystyle \frac{1}{x^m}}+{\displaystyle \frac{1}{x^{2m}}}\right)}^2}+c=f\left(x\right)+c$

$\Rightarrow \frac{1}{2m}\frac{x^{4m}}{{\left(x^{2m}+x^m+1\right)}^2}=f\left(x\right).$