Question Number 88118 by sahnaz last updated on 08/Apr/20
$$\int\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)×\sqrt{\mathrm{x}}} \\ $$
Commented by mathmax by abdo last updated on 08/Apr/20
$${I}\:=\int\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}}}\:\Rightarrow{I}\:=_{{x}={t}^{\mathrm{2}} } \:\:\int\:\:\:\frac{\mathrm{2}{tdt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right){t}}\:=\mathrm{2}\:\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\mathrm{2}{arctan}\left({t}\right)+{C}\:=\mathrm{2}\:{arctan}\left(\sqrt{{x}}\right)\:+{C} \\ $$
Answered by redmiiuser last updated on 08/Apr/20
$$\sqrt{{x}}={z} \\ $$$${dz}=\frac{{dx}}{\mathrm{2}.\sqrt{{x}}} \\ $$$$\int\frac{\mathrm{2}.{dz}}{{z}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\mathrm{2tan}^{−\mathrm{1}} {z}+{c} \\ $$$$=\mathrm{2tan}^{−\mathrm{1}} \sqrt{{x}}+{c} \\ $$$$=\mathrm{2arctan}\:{x}+{c} \\ $$