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

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

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

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

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

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