Question Number 42622 by maxmathsup by imad last updated on 29/Aug/18
$${find}\:\int\:\:{th}\left(\mathrm{2}{x}+\mathrm{1}\right){dx}\: \\ $$
Commented by maxmathsup by imad last updated on 31/Aug/18
$${changement}\:\mathrm{2}{x}+\mathrm{1}\:={t}\:{give} \\ $$$$\int\:{th}\left(\mathrm{2}{x}+\mathrm{1}\right){dx}\:=\:\int\:\:{th}\left({t}\right)\:\frac{{dt}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:{th}\left({t}\right){dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{1}+{e}^{\mathrm{2}{t}} \right)−{t}\:\right\}\:+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{1}+{e}^{\mathrm{4}{x}+\mathrm{2}} \right)−\mathrm{2}{x}−\mathrm{1}\right\}\:+{c}\:. \\ $$