Question Number 227439 by fantastic2 last updated on 28/Jan/26
$$\int\:{x}^{{dx}} \\ $$
Commented by Ghisom_ last updated on 28/Jan/26
$$\:_{\coprod} ^{\Upsilon} {dx}^{\:_{{x}} } \\ $$
Commented by fantastic2 last updated on 28/Jan/26
$${xlnx}−{x}+{C} \\ $$
Commented by Frix last updated on 28/Jan/26
$$\mathrm{Everybody}\:\mathrm{knows}\:\mathrm{that}\:\:_{\coprod} ^{\Upsilon} {dx}^{\:_{{x}} } =\mathrm{42} \\ $$
Answered by Lara2440 last updated on 02/Feb/26
$$\underset{{A}} {\overset{{B}} {\prod}}\:{X}^{\mathrm{d}{X}} ={e}^{\int_{{A}} ^{\:{B}} \:\mathrm{ln}\left({X}\right)\mathrm{d}{X}\:} \:\mathrm{But}\:\int\:{X}^{\mathrm{d}{X}} \:\mathrm{can}'\mathrm{t}\:\mathrm{define} \\ $$$${e}^{\mathrm{d}{X}\centerdot\mathrm{ln}\left({X}\right)} =\underset{\ell=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(\mathrm{ln}\left({X}\right)\mathrm{d}{X}\right)^{\ell} }{\ell!}\approx\mathrm{1}+\mathrm{ln}\left({X}\right)\mathrm{d}{X}+\frac{\mathrm{1}}{\mathrm{2}!}\mathrm{ln}^{\mathrm{2}} \left({X}\right)\mathrm{d}{X}^{\mathrm{2}} +….. \\ $$$$\int\:{e}^{\mathrm{ln}\left({X}\right)\mathrm{d}{X}} \approx\int\:\:\mathrm{1}+\int\:\:\mathrm{ln}\left({X}\right)\mathrm{d}{X}+\underset{=\mathrm{0}} {\underbrace{\frac{\mathrm{1}}{\mathrm{2}!}\int\:\:\mathrm{ln}^{\mathrm{2}} \left({X}\right)\mathrm{d}{X}^{\mathrm{2}} +…..\int\:\mathrm{ln}^{{m}} \left({X}\right)\mathrm{d}{X}^{{m}} }} \\ $$$$\underset{\underset{?????} {\:}} {\underbrace{\int\:\:\underset{\:} {\mathrm{1}}}\:}+\int\:\:\mathrm{ln}\left({X}\right)\mathrm{d}{X} \\ $$$$\:\int\:\:{X}^{\mathrm{d}{X}} \:\:\mathrm{can}'\mathrm{t}\:\mathrm{define}\: \\ $$$$\:\mathrm{but}\:\:\int\:\:{X}^{\mathrm{d}{X}} −\mathrm{1}=\int\:\:\:\mathrm{ln}\left({X}\right)\mathrm{d}{X}={X}\centerdot\mathrm{ln}\left({X}\right)−{X}+{C} \\ $$
Commented by Lara2440 last updated on 02/Feb/26
https://en.wikipedia.org/wiki/Product_integral