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Question Number 49760 by rahul 19 last updated on 10/Dec/18

Commented by rahul 19 last updated on 11/Dec/18

Thank you sir!

$${Thank}\:{you}\:{sir}! \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Dec/18

wait pls trying...

$${wait}\:{pls}\:{trying}... \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Dec/18

∫((x^(5m−1) +2x^(4m−1) )/((x^(2m) +x^m +1)^3 ))dx ∫((x^(5m−1) +2x^(4m−1) )/(x^(6m) (1+x^(−m) +x^(−2m) )^3 ))dx ∫((x^(−m−1) +2x^(−2m−1) )/((1+x^(−m) +x^(−2m) )^3 ))dx t=1+x^(−m) +x^(−2m) (dt/dx)=−mx^(−m−1) −2mx^(−2m−1) ((−dt)/m)=(x^(−m−1) +2x^(−2m−1) )dx ∫((−dt)/(m×t^3 ))=((−1)/m)×(t^(−2) /(−2))=(1/(2m))×(1/((1+x^(−m) +x^(−2m) )^2 ))+c =(1/(2m))×(x^(4m) /((x^(2m) +x^m +1)^2 ))+c

$$\int\frac{{x}^{\mathrm{5}{m}−\mathrm{1}} +\mathrm{2}{x}^{\mathrm{4}{m}−\mathrm{1}} }{\left({x}^{\mathrm{2}{m}} +{x}^{{m}} +\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$$$\int\frac{{x}^{\mathrm{5}{m}−\mathrm{1}} +\mathrm{2}{x}^{\mathrm{4}{m}−\mathrm{1}} }{{x}^{\mathrm{6}{m}} \left(\mathrm{1}+{x}^{−{m}} +{x}^{−\mathrm{2}{m}} \right)^{\mathrm{3}} }{dx} \\ $$$$\int\frac{{x}^{−{m}−\mathrm{1}} +\mathrm{2}{x}^{−\mathrm{2}{m}−\mathrm{1}} }{\left(\mathrm{1}+{x}^{−{m}} +{x}^{−\mathrm{2}{m}} \right)^{\mathrm{3}} }{dx} \\ $$$${t}=\mathrm{1}+{x}^{−{m}} +{x}^{−\mathrm{2}{m}} \\ $$$$\frac{{dt}}{{dx}}=−{mx}^{−{m}−\mathrm{1}} −\mathrm{2}{mx}^{−\mathrm{2}{m}−\mathrm{1}} \\ $$$$\frac{−{dt}}{{m}}=\left({x}^{−{m}−\mathrm{1}} +\mathrm{2}{x}^{−\mathrm{2}{m}−\mathrm{1}} \right){dx} \\ $$$$\int\frac{−{dt}}{{m}×{t}^{\mathrm{3}} }=\frac{−\mathrm{1}}{{m}}×\frac{{t}^{−\mathrm{2}} }{−\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}{m}}×\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{−{m}} +{x}^{−\mathrm{2}{m}} \right)^{\mathrm{2}} }+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{m}}×\frac{{x}^{\mathrm{4}{m}} }{\left({x}^{\mathrm{2}{m}} +{x}^{{m}} +\mathrm{1}\right)^{\mathrm{2}} }+{c} \\ $$$$ \\ $$

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