Question Number 76355 by mathmax by abdo last updated on 26/Dec/19
$${find}\:\int\:\frac{{arctan}\left(\sqrt{\mathrm{1}+{x}}\right)}{\mathrm{2}+{x}}{dx} \\ $$
Answered by john santu last updated on 27/Dec/19
$${let}\:\sqrt{\mathrm{1}+{x}}={tant}\:,\:\mathrm{1}+{x}={tan}^{\mathrm{2}} {t}\:, \\ $$$$\mathrm{2}+{x}={sec}^{\mathrm{2}} {t}\:\:{then}\:{dx}\:=\:\mathrm{2}{tant}\:{sec}^{\mathrm{2}} {t}\:{dt} \\ $$
Answered by john santu last updated on 27/Dec/19
$$=\int\:\frac{\mathrm{2}{t}×{tan}\:{t}\:×{sec}^{\mathrm{2}} {t}\:{dt}}{{sec}^{\mathrm{2}} {t}} \\ $$$$=\:\int\:\mathrm{2}{t}\:{tan}\:{t}\:{dt}\:\left({using}\:{integration}\:{by}\:{part}\right) \\ $$