Question Number 152660 by Tawa11 last updated on 30/Aug/21
$$\int\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}\:\:\:+\:\:\:\mathrm{x}^{\mathrm{3}} }}\:\mathrm{dx} \\ $$
Answered by Olaf_Thorendsen last updated on 31/Aug/21
$$\left(\mathrm{1}+{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{u}^{{n}} \\ $$$${x}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{x}^{\mathrm{3}{n}+\mathrm{1}} \\ $$$$\int{x}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\left(\mathrm{2}{n}\right)!}{\left(\mathrm{3}{n}+\mathrm{2}\right)\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{x}^{\mathrm{3}{n}+\mathrm{2}} +\mathrm{C} \\ $$
Commented by Tawa11 last updated on 31/Aug/21
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$