Question Number 177242 by peter frank last updated on 02/Oct/22
$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{\mathrm{3x}^{\mathrm{16}} +\mathrm{5x}^{\mathrm{14}} }{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }\mathrm{dx} \\ $$$$ \\ $$
Answered by som(math1967) last updated on 02/Oct/22
$$\:\int\frac{\mathrm{3}{x}^{\mathrm{16}} +\mathrm{5}{x}^{\mathrm{14}} }{\left\{{x}^{\mathrm{5}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)\right\}^{\mathrm{4}} }{dx} \\ $$$$\int\frac{{x}^{\mathrm{14}} \left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}\right)}{{x}^{\mathrm{20}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{4}} }{dx} \\ $$$$\int\frac{\frac{\mathrm{3}}{{x}^{\mathrm{4}} }+\frac{\mathrm{5}}{{x}^{\mathrm{6}} }}{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{4}} }{dx} \\ $$$$\:{let}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)={t} \\ $$$$\:\Rightarrow−\left(\frac{\mathrm{3}}{{x}^{\mathrm{4}} }+\frac{\mathrm{5}}{{x}^{\mathrm{6}} }\right){dx}={dt} \\ $$$$\:−\int\frac{{dt}}{{t}^{\mathrm{4}} } \\ $$$$=−\frac{{t}^{−\mathrm{4}+\mathrm{1}} }{−\mathrm{4}+\mathrm{1}}\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{3}} }\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} }\:+{C} \\ $$$$=\frac{{x}^{\mathrm{15}} }{\mathrm{3}\left({x}^{\mathrm{5}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:+{C} \\ $$
Commented by Ar Brandon last updated on 02/Oct/22
wow!
Commented by peter frank last updated on 02/Oct/22
$$\mathrm{thank}\:\mathrm{you} \\ $$