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Question Number 93720 by john santu last updated on 14/May/20

∫ (dx/((x+1)^3 (√(x^2 +2x))))

$$\int\:\frac{\boldsymbol{{dx}}}{\left(\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{3}} \:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}}}\: \\ $$

Commented by john santu last updated on 14/May/20

x^2 +2x = t^(2 ) ⇒ (2x+2) dx = 2t dt dx = ((t dt)/(x+1)) ; (x+1)^2 = t^2 +1⇒(1/((x+1)^2 )) = (1/(t^2 +1)) ∫ (dx/((x+1)^3 (√(x^2 +2x)))) = ∫ ((t dt)/((t^2 +1)^2 .t)) = ∫ (dt/((t^2 +1)^2 )) let t = tan u = ∫ ((sec^2 u du )/(sec^4 u)) = ∫ cos^2 u du = ∫ (1/2)+(1/2)cos 2u du = (1/2)u + (1/2)sin ucos u +c = (1/2)tan^(−1) (t) +(t/(2(t^2 +1))) + c = (1/2)tan^(−1) ((√(x^2 +2x)) )+ ((√(x^2 +2x))/(2(x+1)^2 )) + c

$$\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}\:=\:\boldsymbol{{t}}^{\mathrm{2}\:} \Rightarrow\:\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{2}\right)\:\boldsymbol{{dx}}\:=\:\mathrm{2}\boldsymbol{{t}}\:\boldsymbol{{dt}} \\ $$$$\boldsymbol{{dx}}\:=\:\frac{\boldsymbol{{t}}\:\boldsymbol{{dt}}}{\boldsymbol{{x}}+\mathrm{1}}\:;\:\left(\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{2}} =\:\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\left(\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int\:\frac{\boldsymbol{{dx}}}{\left(\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{3}} \:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}}}\:=\:\int\:\frac{\boldsymbol{{t}}\:\boldsymbol{{dt}}}{\left(\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} .\boldsymbol{{t}}} \\ $$$$=\:\int\:\frac{\boldsymbol{{dt}}}{\left(\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$\mathrm{let}\:\mathrm{t}\:=\:\mathrm{tan}\:\mathrm{u}\: \\ $$$$=\:\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} \boldsymbol{{u}}\:\boldsymbol{{du}}\:}{\mathrm{sec}\:^{\mathrm{4}} \boldsymbol{{u}}}\:=\:\int\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{u}\:\mathrm{du} \\ $$$$=\:\int\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2u}\:\mathrm{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{u}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{ucos}\:\mathrm{u}\:+\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{t}\right)\:+\frac{\mathrm{t}}{\mathrm{2}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)}\:+\:\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:^{−\mathrm{1}} \left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}}\:\right)+\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\:\:+\:\mathrm{c} \\ $$

Commented by mathmax by abdo last updated on 14/May/20

I =∫ (dx/((x+1)^3 (√(x^2 +2x)))) we do the changement (√(x^2 +2x))=t ⇒ x^2 +2x =t^2 ⇒ (2x+2)dx =2tdt ⇒(x+1)dx =t dt ⇒dx =((tdt)/((x+1))) x^2 +2x +1 =t^2 +1 ⇒(x+1)^2 =(t^2 +1) ⇒(dx/((x+1)^3 )) =((tdt)/((t^2 +1))) (dx/((x+1)^3 )) =(dx/((x+1)))×(1/((x+1)^2 )) =((tdt)/((x+1)^2 (x+1)^2 )) =((tdt)/((t^2 +1)^2 )) ⇒ I =∫ ((tdt)/(t(t^2 +1)^2 ))dt =∫ (dt/((t^2 +1)^2 )) =_(t=tanθ) ∫ (((1+tan^2 θ)dθ)/((1+tan^2 θ)^2 )) =∫ (dθ/(1+tan^2 θ)) =∫ cos^2 θ dθ =(1/2)∫(1+cos(2θ)dθ =(1/2)θ +(1/4)sin(2θ) +c =((arctant)/2) +(1/4)×((2t)/(1+t^2 )) +C =((arctant)/2) +(t/(2(1+t^2 ))) +C I=(1/2)arctan((√(x^2 +2x)))+((√(x^2 +2x))/(2(1+x)^2 )) +C

$${I}\:=\int\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}\:\:\:{we}\:{do}\:{the}\:{changement}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}={t}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}\:={t}^{\mathrm{2}} \:\Rightarrow\:\left(\mathrm{2}{x}+\mathrm{2}\right){dx}\:=\mathrm{2}{tdt}\:\Rightarrow\left({x}+\mathrm{1}\right){dx}\:={t}\:{dt}\:\Rightarrow{dx}\:=\frac{{tdt}}{\left({x}+\mathrm{1}\right)} \\ $$$${x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{1}\:={t}^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:=\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\:\Rightarrow\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{{tdt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{{dx}}{\left({x}+\mathrm{1}\right)}×\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{{tdt}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{{tdt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${I}\:=\int\:\:\frac{{tdt}}{{t}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:=\int\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:=_{{t}={tan}\theta} \:\:\:\:\int\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{{d}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:=\int\:{cos}^{\mathrm{2}} \theta\:{d}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right){d}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}\theta\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}\theta\right)\:+{c}\right. \\ $$$$=\frac{{arctant}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+{C}\:=\frac{{arctant}}{\mathrm{2}}\:+\frac{{t}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:+{C} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}\right)+\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}{\mathrm{2}\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:+{C} \\ $$

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