Question Number 197767 by AR19 last updated on 28/Sep/23
Commented by Frix last updated on 28/Sep/23
$$\int\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}{dx}={x}\:_{\mathrm{2}} {F}_{\mathrm{1}} \:\left(−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{3}};\:\frac{\mathrm{4}}{\mathrm{3}};\:−{x}^{\mathrm{3}} \right)\:+{C} \\ $$