Question Number 43159 by MASANJA J last updated on 07/Sep/18
Answered by alex041103 last updated on 08/Sep/18
$${For}\:\int\frac{{dx}}{\:\sqrt{{x}+\mathrm{15}}}\:: \\ $$$$\int\frac{{dx}}{\:\sqrt{{x}+\mathrm{15}}}\:=\:\int\frac{{d}\left({x}+\mathrm{15}\right)}{\:\sqrt{{x}+\mathrm{15}}}=\int{u}^{−\mathrm{1}/\mathrm{2}} {du}= \\ $$$$=\mathrm{2}{u}^{\mathrm{1}/\mathrm{2}} +{C}=\mathrm{2}\sqrt{{x}+\mathrm{15}}\:+{C} \\ $$$$ \\ $$$${For}\:\int\frac{{ln}\left({x}\right){dx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:{see}\:{Q}.\mathrm{43191} \\ $$