If m is a non-zero number and int frac{x^{5 m | vedclass

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(B) Given integral: $I = \int \frac{x^{5 m-1}+2 x^{4 m-1}}{\left(x^{2 m}+x^{m}+1\right)^{3}} d x$Divide numerator and denominator by $x^{6m}$:$I = \in

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$\frac{x^{4m}}{2m(x^{2m}+x^m+1)^2}$

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(B) Given integral: $I = \int \frac{x^{5 m-1}+2 x^{4 m-1}}{\left(x^{2 m}+x^{m}+1\right)^{3}} d x$Divide numerator and denominator by $x^{6m}$:$I = \int \frac{x^{5 m-1}+2 x^{4 m-1}}{x^{6m}\left(x^{-2m}+x^{-m}+1\right)^{3}} d x$$I = \int \frac{x^{-m-1}+2 x^{-2 m-1}}{\left(1+x^{-m}+x^{-2 m}\right)^{3}} d x$Let $t = 1+x^{-m}+x^{-2 m}$.Then $dt = (-m x^{-m-1} - 2m x^{-2m-1}) dx = -m(x^{-m-1} + 2x^{-2m-1}) dx$.So,$(x^{-m-1} + 2x^{-2m-1}) dx = -\frac{dt}{m}$.Substituting into the integral:$I = \int \frac{-dt/m}{t^3} = -\frac{1}{m} \int t^{-3} dt = -\frac{1}{m} \left( \frac{t^{-2}}{-2} \right) + C = \frac{1}{2mt^2} + C$.Substituting $t$ back:$I = \frac{1}{2m(1+x^{-m}+x^{-2m})^2} + C = \frac{1}{2m(\frac{x^{2m}+x^m+1}{x^{2m}})^2} + C = \frac{x^{4m}}{2m(x^{2m}+x^m+1)^2} + C$.Thus,$f(x) = \frac{x^{4m}}{2m(x^{2m}+x^m+1)^2}$.

If $m$ is a non-zero number and $\int \frac{x^{5 m-1}+2 x^{4 m-1}}{\left(x^{2 m}+x^{m}+1\right)^{3}} d x=f(x)+c$,then $f(x)$ is

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