x-x-1-x-2-dx-

Question Number 84396 by Aditya789 last updated on 12/Mar/20

$$\int\frac{\mathrm{x}\sqrt{\mathrm{x}+\mathrm{1}}}{\mathrm{x}+\mathrm{2}}\mathrm{dx} \\ $$

Commented by jagoll last updated on 12/Mar/20

let (√(x+1 )) = u ⇒ x=u^2 −1 ⇒∫ (( (u^2 −1)u (2u du))/(u^2 +1)) = ∫ (( 2u^4 −2u^2 )/(u^2 +1)) du = ∫ (2u^2 −4)du + ∫ (4/(u^2 +1)) du = (2/3)u^3 −4u+4 tan^(−1) (u) + c = (2/3)u (u^2 −6)+4tan^(−1) (u)+c = (2/3)(x−5)(√(x+1)) + 4tan^(−1) ((√(x+1))) +c

$$\mathrm{let}\:\sqrt{\mathrm{x}+\mathrm{1}\:}\:=\:\mathrm{u}\:\Rightarrow\:\mathrm{x}=\mathrm{u}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\int\:\frac{\:\:\left(\mathrm{u}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{u}\:\left(\mathrm{2u}\:\mathrm{du}\right)}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}\:=\: \\ $$$$\int\:\frac{\:\mathrm{2u}^{\mathrm{4}} −\mathrm{2u}^{\mathrm{2}} }{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{du}\:= \\ $$$$\int\:\left(\mathrm{2u}^{\mathrm{2}} −\mathrm{4}\right)\mathrm{du}\:+\:\int\:\frac{\mathrm{4}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{du}\:=\: \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\mathrm{u}^{\mathrm{3}} −\mathrm{4u}+\mathrm{4}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{u}\right)\:+\:\mathrm{c}\:= \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\mathrm{u}\:\left(\mathrm{u}^{\mathrm{2}} −\mathrm{6}\right)+\mathrm{4tan}^{−\mathrm{1}} \left(\mathrm{u}\right)+\mathrm{c}\:= \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{x}−\mathrm{5}\right)\sqrt{\mathrm{x}+\mathrm{1}}\:+\:\mathrm{4tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{x}+\mathrm{1}}\right)\:+\mathrm{c} \\ $$

Commented by mathmax by abdo last updated on 12/Mar/20

I =∫ ((x(√(x+1)))/(x+2))dx changement (√(x+1))=tgive x+1 =t^2 ⇒ I =∫ (((t^2 −1).t)/(t^(2 ) +1))(2t)dt =2∫ ((t^4 −t^2 )/(t^2 +1))dt =2 ∫ ((t^2 (t^2 +1)−2t^2 )/(t^2 +1))dt =2{ ∫ t^2 dt −2∫ ((t^2 +1−1)/(t^2 +1))dt} =2{(t^3 /3) −2t + 2arctan(t)} +C =(2/3)((√(x+1)))^3 −4 (√(x+1)) +4arctan((√(x+1))) +C

$${I}\:=\int\:\:\frac{{x}\sqrt{{x}+\mathrm{1}}}{{x}+\mathrm{2}}{dx}\:\:{changement}\:\sqrt{{x}+\mathrm{1}}={tgive}\:{x}+\mathrm{1}\:={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int\:\:\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right).{t}}{{t}^{\mathrm{2}\:} +\mathrm{1}}\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\int\:\:\frac{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$$=\mathrm{2}\:\int\:\:\frac{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} \:+\mathrm{1}\right)−\mathrm{2}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:=\mathrm{2}\left\{\:\int\:{t}^{\mathrm{2}} \:{dt}\:−\mathrm{2}\int\:\:\frac{{t}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\right\} \\ $$$$=\mathrm{2}\left\{\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\:−\mathrm{2}{t}\:+\:\mathrm{2}{arctan}\left({t}\right)\right\}\:+{C} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\sqrt{{x}+\mathrm{1}}\right)^{\mathrm{3}} −\mathrm{4}\:\sqrt{{x}+\mathrm{1}}\:+\mathrm{4}{arctan}\left(\sqrt{{x}+\mathrm{1}}\right)\:+{C} \\ $$

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