Question Number 149989 by ZiYangLee last updated on 08/Aug/21
$$\mathrm{By}\:\mathrm{subs}\:{u}^{\mathrm{2}} =\mathrm{4}+{x},\:\mathrm{evaluate}\:\int\:\frac{\sqrt{\mathrm{4}+{x}}}{{x}}\:{dx} \\ $$
Answered by puissant last updated on 08/Aug/21
$$\int\frac{\sqrt{\mathrm{4}+{x}}}{{x}}{dx}={Q} \\ $$$${u}=\sqrt{\mathrm{4}+{x}}\:\rightarrow\:{u}^{\mathrm{2}} =\mathrm{4}+{x}\:\rightarrow\:{x}={u}^{\mathrm{2}} −\mathrm{4} \\ $$$${dx}=\mathrm{2}{udu} \\ $$$${Q}=\int\frac{{u}}{{u}^{\mathrm{2}} −\mathrm{4}}\mathrm{2}{udu} \\ $$$$=\mathrm{2}\int\frac{{u}^{\mathrm{2}} }{{u}^{\mathrm{2}} −\mathrm{4}}{du} \\ $$$$=\mathrm{2}\left(\int\frac{{u}^{\mathrm{2}} −\mathrm{4}}{{u}^{\mathrm{2}} −\mathrm{4}}{du}+\mathrm{4}\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} −\mathrm{4}}{du}\right) \\ $$$$=\mathrm{2}{u}+\mathrm{8}\int\left(\frac{{a}}{{u}−\mathrm{2}}+\frac{{b}}{{u}+\mathrm{2}}\right){du} \\ $$$${we}\:{have}\:\:: \\ $$$$\frac{\mathrm{1}}{\left({u}−\mathrm{2}\right)\left({u}+\mathrm{2}\right)}{au}+\mathrm{2}{a}+{ub}−\mathrm{2}{b}=\frac{\mathrm{1}}{\left({u}−\mathrm{2}\right)\left({u}+\mathrm{2}\right)} \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{2}{a}−\mathrm{2}{b}=\mathrm{1}}\\{{a}+{b}=\mathrm{0}}\end{cases}\Rightarrow\:\begin{cases}{\mathrm{2}{a}+\mathrm{2}{b}=\mathrm{0}}\\{\mathrm{2}{a}−\mathrm{2}{b}=\mathrm{1}}\end{cases} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{4}}\:{et}\:{b}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{Q}=\mathrm{2}{u}+\mathrm{8}\int\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{{u}−\mathrm{2}}{du}−\mathrm{8}\int\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{{u}+\mathrm{2}}{du} \\ $$$$\Rightarrow\:{Q}=\mathrm{2}{u}+\mathrm{2}{ln}\left(\frac{{u}−\mathrm{2}}{{u}+\mathrm{2}}\right)+{C} \\ $$$${soit}\:{Q}=\mathrm{2}\sqrt{\mathrm{4}+{x}}+\mathrm{2}{ln}\left(\frac{\sqrt{\mathrm{4}+{x}}−\mathrm{2}}{\:\sqrt{\mathrm{4}+{x}}+\mathrm{2}}\right)+{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:…..{Le}\:{puissant}…. \\ $$
Commented by peter frank last updated on 08/Aug/21
$${thank}\:{you} \\ $$