If m is a non-zero number and ∫ x ^ 5m - 1 + 2 x ^ 4m - 1 ( x ^ 2m...

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

$⇒ ∫ \dfrac{{x}^{5m - 1} + 2{x}^{4m - 1}}{{x}^{6m}{\left(1 + {x}^{- m} + {x}^{- 2m}\right)}^{3}}dx = f\left(x\right) + c$

$⇒ ∫ \dfrac{{x}^{- m - 1} + 2{x}^{- 2m - 1}}{{\left(1 + {x}^{- m} + {x}^{- 2m}\right)}^{3}}dx = f\left(x\right) + c$.

If, $1 + {x}^{- m} + {x}^{- 2m} = t$

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

$∴ ∫ \dfrac{- dt}{m{t}^{3}} = f\left(x\right) + c$

$⇒ - \dfrac{1}{m} ∫ {t}^{- 3}dt = f\left(x\right) + c$

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

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

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

$⇒ \dfrac{1}{2m{\left(1 + \dfrac{1}{{x}^{m}} + \dfrac{1}{{x}^{2m}}\right)}^{2}} + c = f\left(x\right) + c$

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