Question Number 198403 by Mingma last updated on 19/Oct/23
Answered by MM42 last updated on 19/Oct/23
$$\sqrt{{x}+\mathrm{2}}={u}\Rightarrow{x}={u}^{\mathrm{2}} −\mathrm{2}\Rightarrow{dx}=\mathrm{2}{udu} \\ $$$$\Rightarrow\int\:\frac{\mathrm{2}{udu}}{{u}^{\mathrm{2}} −{u}−\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{3}}\int\left(\frac{\mathrm{2}}{{u}−\mathrm{2}}+\frac{\mathrm{1}}{{u}+\mathrm{1}}\right){du} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}{ln}\left({u}−\mathrm{2}\right)+\frac{\mathrm{2}}{\mathrm{3}}{ln}\left({u}+\mathrm{1}\right)+{c} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}{ln}\left(\sqrt{{x}+\mathrm{2}}−\mathrm{2}\right)+\frac{\mathrm{2}}{\mathrm{3}}{ln}\left(\sqrt{{x}+\mathrm{2}}+\mathrm{1}\right)+{c}\:\:\checkmark \\ $$$$ \\ $$