Question Number 7792 by Tawakalitu. last updated on 15/Sep/16
Commented by FilupSmith last updated on 16/Sep/16
$$\left(\mathrm{3}\right)\:\:\int_{\mathrm{1}} ^{\:\mathrm{3}} {x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{{x}=\mathrm{1}} ^{\:{x}=\mathrm{3}} {x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$${u}={x}^{\mathrm{2}} −\mathrm{1} \\ $$$${du}=\mathrm{2}{xdx} \\ $$$$\int_{\mathrm{1}} ^{\:\mathrm{3}} {x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{{x}=\mathrm{1}} ^{\:{x}=\mathrm{3}} \sqrt{{u}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{{x}=\mathrm{1}} ^{\:{x}=\mathrm{3}} \left({u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left({u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \frac{\mathrm{2}}{\mathrm{3}}\right]_{{x}=\mathrm{1}} ^{{x}=\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{{x}=\mathrm{1}} ^{{x}=\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\left(\mathrm{9}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{1}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\left(\mathrm{9}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{1}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(^{\:\mathrm{2}} \sqrt{\mathrm{8}^{\mathrm{3}} }\right)=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{16}\sqrt{\mathrm{2}} \\ $$$$=\frac{\mathrm{16}}{\mathrm{3}}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$−−−−−−−−−−−−−−−−−−−−− \\ $$$$ \\ $$$$\left(\mathrm{4}\right)\:\int\:\frac{{e}^{{x}} −{e}^{−{x}} }{{e}^{{x}} +{e}^{−{x}} } \\ $$$${e}^{{x}} +{e}^{−{x}} ={u} \\ $$$${du}={e}^{{x}} −{e}^{{x}} \:{dx} \\ $$$$=\int\frac{\mathrm{1}}{{u}}{du} \\ $$$$=\mathrm{ln}\left({e}^{{x}} +{e}^{−{x}} \right)+{c} \\ $$$$ \\ $$$$−−−−−−−−−−−−−−−−−−−−− \\ $$$$ \\ $$$$\left(\mathrm{5}\right)\:\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\left(\mathrm{2}{x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{ln}\left(\mathrm{7}\right)−\mathrm{ln}\left(\mathrm{1}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{7}\right) \\ $$$$ \\ $$$${or} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}{dx} \\ $$$${u}=\mathrm{2}{x}+\mathrm{1} \\ $$$${du}=\mathrm{2}{dx}\:\:\Rightarrow\:\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}{du} \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{1}}{{u}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){du} \\ $$$${same}\:{as}\:{above} \\ $$
Commented by Tawakalitu. last updated on 16/Sep/16
$${Thank}\:{you}\:{sir}.\:{for}\:{your}\:{help}. \\ $$
Commented by Tawakalitu. last updated on 16/Sep/16
$${Thanks}\:{so}\:{much}\:{sir},\:{i}\:{really}\:{appreciate}. \\ $$