Question Number 164171 by mkam last updated on 15/Jan/22
$$\int\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{10}} } \\ $$
Answered by Ar Brandon last updated on 15/Jan/22
$$\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{10}} }=\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{x}^{\mathrm{10}} \right)^{{n}} {dx}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{1}\right)_{{n}} \left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{10}{n}+\mathrm{1}} }{{n}!\left(\mathrm{10}{n}+\mathrm{1}\right)} \\ $$$$=\frac{{x}}{\mathrm{10}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{1}\right)_{{n}} \left(−{x}^{\mathrm{10}} \right)^{{n}} }{{n}!\left({n}+\frac{\mathrm{1}}{\mathrm{10}}\right)}=\frac{{x}}{\mathrm{10}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{1}\right)_{{n}} \Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{10}}\right)}{{n}!\Gamma\left({n}+\frac{\mathrm{11}}{\mathrm{10}}\right)}\left(−{x}^{\mathrm{10}} \right)^{{n}} \\ $$$$=\frac{{x}}{\mathrm{100}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{1}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{10}}\right)_{{n}} }{{n}!\left(\frac{\mathrm{11}}{\mathrm{10}}\right)_{{n}} }\left(−{x}^{\mathrm{10}} \right)^{{n}} =\frac{{x}}{\mathrm{100}}\:\underset{\mathrm{2}} {\:}{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{10}},\:\mathrm{1};\:\frac{\mathrm{11}}{\mathrm{10}};\:−{x}^{\mathrm{10}} \right)+{C} \\ $$