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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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