Question Number 132647 by aurpeyz last updated on 15/Feb/21
Answered by Dwaipayan Shikari last updated on 15/Feb/21
$$\int\frac{\mathrm{1}}{\:\sqrt{{x}}\left(\sqrt{{x}}+\mathrm{7}\right)}{dx}\:\:\:\:\:\:\:\:{x}={t}^{\mathrm{2}} \\ $$$$=\mathrm{2}\int\frac{\mathrm{1}}{{t}+\mathrm{7}}{dt}=\mathrm{2}{log}\left({t}+\mathrm{7}\right)+{C}={log}\left(\sqrt{{x}}+\mathrm{7}\right)^{\mathrm{2}} +{C} \\ $$
Answered by EDWIN88 last updated on 15/Feb/21
$$\mathrm{let}\:\sqrt{\mathrm{x}}\:+\:\mathrm{7}\:=\:\mathrm{u}\:\mathrm{and}\:\mathrm{x}\:=\:\left(\mathrm{u}−\mathrm{7}\right)^{\mathrm{2}} \\ $$$$\mathrm{I}=\:\int\:\frac{\mathrm{2}\left(\mathrm{u}−\mathrm{7}\right)}{\left(\mathrm{u}−\mathrm{7}\right)\mathrm{u}}\:\mathrm{du}\:=\:\int\:\frac{\mathrm{2}}{\mathrm{u}}\:\mathrm{du}\:=\:\mathrm{2ln}\:\mathrm{u}\:+\:\mathrm{c}\: \\ $$$$\mathrm{I}=\:\mathrm{2ln}\:\left(\sqrt{\mathrm{x}}\:+\:\mathrm{7}\right)\:+\:\mathrm{c}\: \\ $$