Question Number 8763 by tawakalitu last updated on 26/Oct/16
$$\int\mathrm{x}\sqrt{\mathrm{3x}\:+\:\mathrm{1}}\:\:\mathrm{dx} \\ $$
Commented by FilupSmith last updated on 26/Oct/16
$${u}=\mathrm{3}{x}+\mathrm{2}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{3}}\left({u}−\mathrm{2}\right) \\ $$$${du}=\mathrm{3}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int\mathrm{3}{x}\sqrt{\mathrm{3}{x}+\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{3}}\int\left({u}−\mathrm{2}\right)\sqrt{{u}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\int\left({u}−\mathrm{2}\right)\sqrt{{u}}{du} \\ $$$$\mathrm{continue} \\ $$
Commented by tawakalitu last updated on 26/Oct/16
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$
Answered by sou1618 last updated on 26/Oct/16
$${I}=\int{x}\sqrt{\mathrm{3}{x}+\mathrm{1}}{dx} \\ $$$${t}=\mathrm{3}{x}+\mathrm{1}\Rightarrow{x}=\frac{{t}−\mathrm{1}}{\mathrm{3}} \\ $$$${dx}=\frac{\mathrm{1}}{\mathrm{3}}{dt} \\ $$$${I}=\int\frac{{t}−\mathrm{1}}{\mathrm{3}}\sqrt{{t}}×\frac{\mathrm{1}}{\mathrm{3}}{dt} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{9}}\int{t}^{\mathrm{3}/\mathrm{2}} −{t}^{\mathrm{1}/\mathrm{2}} {dt} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{\mathrm{2}}{\mathrm{5}}{t}^{\mathrm{5}/\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}/\mathrm{2}} \right)+{C} \\ $$$$\:\:=\frac{\mathrm{2}}{\mathrm{45}}{t}^{\mathrm{5}/\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{27}}{t}^{\mathrm{3}/\mathrm{2}} +{C} \\ $$$$\:\:=\frac{\mathrm{2}}{\mathrm{45}}\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{5}/\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{27}}\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} +{C} \\ $$$$ \\ $$
Commented by tawakalitu last updated on 26/Oct/16
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$$$ \\ $$