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x-2x-1-dx-




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

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