Question Number 49133 by AdqhK ÐQeQqQ last updated on 03/Dec/18
$$\int\frac{\mathrm{1}}{{x}^{\mathrm{5}} +\mathrm{1}}{dx}=?? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18
$${pls}\:{mention}\:{the}\:{source}\:{of}\:{this}\:{intregal}… \\ $$
Answered by MJS last updated on 04/Dec/18
$$\frac{\mathrm{1}}{{x}^{\mathrm{5}} +\mathrm{1}}=\frac{{A}}{{x}+\mathrm{1}}+\frac{{Bx}+{C}}{{x}^{\mathrm{2}} −\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x}+\mathrm{1}}+\frac{{Dx}+{E}}{{x}^{\mathrm{2}} −\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}+\mathrm{1}} \\ $$$$\mathrm{with} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{5}};\:{B}=−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{10}};\:{C}={E}=\frac{\mathrm{2}}{\mathrm{5}};\:{D}=−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$$ \\ $$$$\mathrm{or} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{5}} +\mathrm{1}}=\frac{{A}}{{x}−\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{5}}} }+\frac{{B}}{{x}−\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}\pi}{\mathrm{5}}} }+\frac{{C}}{{x}+\mathrm{1}}+\frac{{D}}{{x}−\mathrm{e}^{\mathrm{i}\frac{\mathrm{7}\pi}{\mathrm{5}}} }+\frac{{E}}{{x}−\mathrm{e}^{\mathrm{i}\frac{\mathrm{9}\pi}{\mathrm{5}}} } \\ $$$$\mathrm{with} \\ $$$${A}=−\mathrm{1};\:{B}=\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}\pi}{\mathrm{5}}} ;\:{C}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{5}}} ;\:{D}=\mathrm{e}^{\mathrm{i}\frac{\mathrm{9}\pi}{\mathrm{5}}} ;\:{E}=\mathrm{e}^{\mathrm{i}\frac{\mathrm{7}\pi}{\mathrm{5}}} \\ $$$$ \\ $$$$…\mathrm{then}\:\mathrm{use}\:\mathrm{standard}\:\mathrm{formulas}\:\mathrm{for} \\ $$$$\int\frac{{a}}{{x}−{b}}{dx}\:\mathrm{and}\:\int\frac{{ax}+{b}}{{x}^{\mathrm{2}} +{cx}+{d}}{dx} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18
$${excellent}\:{sir}… \\ $$
Commented by MJS last updated on 04/Dec/18
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by AdqhK ÐQeQqQ last updated on 04/Dec/18
$${thanku}\:{very}\:{much}\:{sir} \\ $$