Question Number 7814 by Tawakalitu. last updated on 16/Sep/16
$$\int{x}\sqrt{\mathrm{2}{x}\:+\:\mathrm{1}}\:{dx}\: \\ $$
Commented by l-becker last updated on 17/Sep/16
$$ \\ $$$$\frac{{x}}{{dx}}\left(\sqrt{\mathrm{2}{x}+\mathrm{1}}\:\right) \\ $$$$\frac{{x}}{{dx}}\int\sqrt{\mathrm{4}{x}+\mathrm{1}}\:+{C} \\ $$$$\int{x}\sqrt{\mathrm{2}{x}+\mathrm{1}}\:{dx}=\frac{{x}}{{dx}}\int\sqrt{\mathrm{4}{x}+\mathrm{1}}\:+{C} \\ $$$$ \\ $$
Commented by ridwan balatif last updated on 17/Sep/16
$${derivative}\:\:\:\:\:\:\:\mid\:\:\:{integral} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$−−−−−−−−−−−−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\:\:\:\:\:\:\:.\left(+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{2}}{\mathrm{5}}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{15}}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:\:\:\:\:\:\:\:\:.\left(−\mathrm{1}\right) \\ $$$$\int{x}\sqrt{\mathrm{2}{x}+\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{3}}{x}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{15}}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} +{C} \\ $$$$\int{x}\sqrt{\mathrm{2}{x}+\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{15}}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{3}{x}−\mathrm{1}\right)+{C} \\ $$$${i}\:{use}\:{tanzalin}\:{ways} \\ $$
Commented by FilupSmith last updated on 17/Sep/16
$${u}=\mathrm{2}{x}+\mathrm{1}\:\:\:\Rightarrow\:\:\:{x}=\frac{{u}−\mathrm{1}}{\mathrm{2}} \\ $$$${du}=\mathrm{2}{dx} \\ $$$$\int{x}\sqrt{\mathrm{2}{x}+\mathrm{1}}{dx}=\int\:\frac{{u}−\mathrm{1}}{\mathrm{2}}\sqrt{{u}}×\frac{\mathrm{1}}{\mathrm{2}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left({u}−\mathrm{1}\right)\sqrt{{u}}\:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left({u}^{\mathrm{3}/\mathrm{2}} −{u}^{\mathrm{1}/\mathrm{2}} \right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{2}}{\mathrm{5}}{u}^{\mathrm{5}/\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}/\mathrm{2}} \right)+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{2}}{\mathrm{5}}{u}^{\mathrm{5}/\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}/\mathrm{2}} \right)+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{20}}{u}^{\mathrm{5}/\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{15}}{u}^{\mathrm{3}/\mathrm{2}} +{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{5}/\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{15}}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} +{c} \\ $$$${working} \\ $$