Question Number 191567 by leandrosriv02 last updated on 26/Apr/23
Answered by Frix last updated on 26/Apr/23
$$\int\frac{\sqrt{{x}}−\mathrm{1}}{{x}^{\pi} }{dx}=\int{x}^{\frac{\mathrm{1}}{\mathrm{2}}−\pi} {dx}−\int{x}^{−\pi} {dx} \\ $$$$\mathrm{Which}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{using}\:\int{x}^{{r}} {dx}=\frac{{x}^{{r}+\mathrm{1}} }{{r}+\mathrm{1}} \\ $$$$\mathrm{but}\:\mathrm{within}\:\left[\mathrm{0},\:\mathrm{2}\right]\:\mathrm{the}\:\mathrm{integral}\:\mathrm{does}\:\mathrm{not} \\ $$$$\mathrm{converge}. \\ $$