Question Number 84123 by Roland Mbunwe last updated on 09/Mar/20
$$\int\frac{\mathrm{5}−\boldsymbol{{x}}}{\mathrm{1}+\sqrt{\left(\boldsymbol{{x}}−\mathrm{4}\right)}} \\ $$
Commented by mathmax by abdo last updated on 09/Mar/20
$${I}=\int\:\frac{\mathrm{5}−{x}}{\mathrm{1}+\sqrt{{x}−\mathrm{4}}}{dx}\:{we}\:{do}\:{the}\:{changement}\:\sqrt{{x}−\mathrm{4}}={t}\:\Rightarrow{x}−\mathrm{4}={t}^{\mathrm{2}} \\ $$$${I}\:=\int\:\frac{\mathrm{5}−\left(\mathrm{4}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}}\:\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int\:\:\frac{{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}}{dt} \\ $$$$=\mathrm{2}\int\:\frac{{t}\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}{dt}\:=\mathrm{2}\int\:\left({t}−{t}^{\mathrm{2}} \right){dt}\:=\mathrm{2}\left(\frac{{t}^{\mathrm{2}} }{\mathrm{2}}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\right)+{C} \\ $$$$={t}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} \:+{C}\:={x}−\mathrm{4}\:−\frac{\mathrm{2}}{\mathrm{3}}\left({x}−\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{C} \\ $$