Question Number 49760 by rahul 19 last updated on 10/Dec/18
Commented by rahul 19 last updated on 11/Dec/18
$${Thank}\:{you}\:{sir}! \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Dec/18
$${wait}\:{pls}\:{trying}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Dec/18
$$\int\frac{{x}^{\mathrm{5}{m}−\mathrm{1}} +\mathrm{2}{x}^{\mathrm{4}{m}−\mathrm{1}} }{\left({x}^{\mathrm{2}{m}} +{x}^{{m}} +\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$$$\int\frac{{x}^{\mathrm{5}{m}−\mathrm{1}} +\mathrm{2}{x}^{\mathrm{4}{m}−\mathrm{1}} }{{x}^{\mathrm{6}{m}} \left(\mathrm{1}+{x}^{−{m}} +{x}^{−\mathrm{2}{m}} \right)^{\mathrm{3}} }{dx} \\ $$$$\int\frac{{x}^{−{m}−\mathrm{1}} +\mathrm{2}{x}^{−\mathrm{2}{m}−\mathrm{1}} }{\left(\mathrm{1}+{x}^{−{m}} +{x}^{−\mathrm{2}{m}} \right)^{\mathrm{3}} }{dx} \\ $$$${t}=\mathrm{1}+{x}^{−{m}} +{x}^{−\mathrm{2}{m}} \\ $$$$\frac{{dt}}{{dx}}=−{mx}^{−{m}−\mathrm{1}} −\mathrm{2}{mx}^{−\mathrm{2}{m}−\mathrm{1}} \\ $$$$\frac{−{dt}}{{m}}=\left({x}^{−{m}−\mathrm{1}} +\mathrm{2}{x}^{−\mathrm{2}{m}−\mathrm{1}} \right){dx} \\ $$$$\int\frac{−{dt}}{{m}×{t}^{\mathrm{3}} }=\frac{−\mathrm{1}}{{m}}×\frac{{t}^{−\mathrm{2}} }{−\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}{m}}×\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{−{m}} +{x}^{−\mathrm{2}{m}} \right)^{\mathrm{2}} }+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{m}}×\frac{{x}^{\mathrm{4}{m}} }{\left({x}^{\mathrm{2}{m}} +{x}^{{m}} +\mathrm{1}\right)^{\mathrm{2}} }+{c} \\ $$$$ \\ $$