Question Number 7961 by tawakalitu last updated on 25/Sep/16
$$\int\frac{\mathrm{5}\:−\:{x}}{\mathrm{1}\:+\:\sqrt{{x}\:−\:\mathrm{4}}}\:{dx} \\ $$
Answered by Yozzia last updated on 25/Sep/16
$${I}=\int\frac{\mathrm{5}−{x}}{\mathrm{1}+\sqrt{{x}−\mathrm{4}}}{dx} \\ $$$${Let}\:{t}=\sqrt{{x}−\mathrm{4}}\Rightarrow{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{4}\right)^{−\mathrm{1}/\mathrm{2}} {dx} \\ $$$${dx}=\mathrm{2}\left({x}−\mathrm{4}\right)^{\mathrm{1}/\mathrm{2}} {dt}=\mathrm{2}{tdt} \\ $$$${x}={t}^{\mathrm{2}} +\mathrm{4} \\ $$$$\therefore{I}=\int\frac{\mathrm{5}−\mathrm{4}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}}×\mathrm{2}{tdt}=\int\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}}{dt} \\ $$$${I}=\int\frac{\mathrm{2}{t}\left(\mathrm{1}+{t}\right)\left(\mathrm{1}−{t}\right)}{\mathrm{1}+{t}}{dt}=\int\mathrm{2}{t}\left(\mathrm{1}−{t}\right){dt} \\ $$$${I}=\int\left(\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{2}} \right){dt}={t}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} +{c} \\ $$$${I}={x}−\mathrm{4}−\frac{\mathrm{2}}{\mathrm{3}}\left({x}−\mathrm{4}\right)^{\mathrm{3}/\mathrm{2}} +{c}={x}−\frac{\mathrm{2}}{\mathrm{3}}\left({x}−\mathrm{4}\right)^{\mathrm{3}/\mathrm{2}} +{D} \\ $$
Commented by tawakalitu last updated on 25/Sep/16
$${Thanks}\:{so}\:{much}\:{sir}. \\ $$