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Question Number 219193 by fantastic last updated on 20/Apr/25
∫(√((x+1)/(x+2))) .(1/(x+3)) dx=?
$$\int\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}\:.\frac{\mathrm{1}}{{x}+\mathrm{3}}\:{dx}=? \\ $$
Answered by aleks041103 last updated on 20/Apr/25
((x+1)/(x+2))=u=1−(1/(x+2)) ⇒x=(1/(1−u))−2=((2u−1)/(1−u)) ⇒dx=((2(1−u)−(−1)(2u−1))/((1−u)^2 ))du= =((2−2u+2u−1)/((1−u)^2 ))du=(du/((1−u)^2 )) (1/(x+3))=(1/(1+(1/(1−u))))=((1−u)/(2−u))=((u−1)/(u−2)) ⇒I=∫(√(((x+1)/(x+2)) ))(dx/(x+3)) = ∫(√u) ((u−1)/(u−2)) (du/((u−1)^2 ))= =∫ (((√u)du)/((u−1)(u−2))) (1/((u−1)(u−2))) = (1/(u−2))−(1/(u−1)) ⇒ I=∫ (((√u)du)/(u−2)) − ∫ (((√u)du)/(u−1)) ∫ (((√u)du)/(u−a)) = ∫((2u)/(u−a)) (du/(2(√u))) = 2J_a z=(√u)=(√((x+1)/(x+2))) ⇒J_a = ∫(z^2 /(z^2 −a))dz=∫1+(a/(z^2 −a))dz= =z+(√a)∫((d(z/(√a)))/((z/(√a))^2 −1))= =z+((√a)/2) ln∣((z−(√a))/(z+(√a)))∣ ⇒I=2(J_2 −J_1 )= =(√2)ln∣((z−(√2))/(z+(√2)))∣−ln∣((z−1)/(z+1))∣ ⇒∫(√((x+1)/(x+2))) (dx/(x+3)) = (√2) ln∣(((√((x+1)/(x+2)))−(√2))/( (√((x+1)/(x+2)))+(√2)))∣−ln∣(((√((x+1)/(x+2)))−1)/( (√((x+1)/(x+2)))+1))∣+C
$$\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}={u}=\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{1}−{u}}−\mathrm{2}=\frac{\mathrm{2}{u}−\mathrm{1}}{\mathrm{1}−{u}} \\ $$$$\Rightarrow{dx}=\frac{\mathrm{2}\left(\mathrm{1}−{u}\right)−\left(−\mathrm{1}\right)\left(\mathrm{2}{u}−\mathrm{1}\right)}{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} }{du}= \\ $$$$=\frac{\mathrm{2}−\mathrm{2}{u}+\mathrm{2}{u}−\mathrm{1}}{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} }{du}=\frac{{du}}{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{x}+\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{u}}}=\frac{\mathrm{1}−{u}}{\mathrm{2}−{u}}=\frac{{u}−\mathrm{1}}{{u}−\mathrm{2}} \\ $$$$\Rightarrow{I}=\int\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\:}\frac{{dx}}{{x}+\mathrm{3}}\:=\:\int\sqrt{{u}}\:\frac{{u}−\mathrm{1}}{{u}−\mathrm{2}}\:\frac{{du}}{\left({u}−\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$=\int\:\frac{\sqrt{{u}}{du}}{\left({u}−\mathrm{1}\right)\left({u}−\mathrm{2}\right)} \\ $$$$\frac{\mathrm{1}}{\left({u}−\mathrm{1}\right)\left({u}−\mathrm{2}\right)}\:=\:\frac{\mathrm{1}}{{u}−\mathrm{2}}−\frac{\mathrm{1}}{{u}−\mathrm{1}} \\ $$$$\Rightarrow\:{I}=\int\:\frac{\sqrt{{u}}{du}}{{u}−\mathrm{2}}\:−\:\int\:\frac{\sqrt{{u}}{du}}{{u}−\mathrm{1}} \\ $$$$\int\:\frac{\sqrt{{u}}{du}}{{u}−{a}}\:=\:\int\frac{\mathrm{2}{u}}{{u}−{a}}\:\frac{{du}}{\mathrm{2}\sqrt{{u}}}\:=\:\mathrm{2}{J}_{{a}} \\ $$$${z}=\sqrt{{u}}=\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}} \\ $$$$\Rightarrow{J}_{{a}} =\:\int\frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{2}} −{a}}{dz}=\int\mathrm{1}+\frac{{a}}{{z}^{\mathrm{2}} −{a}}{dz}= \\ $$$$={z}+\sqrt{{a}}\int\frac{{d}\left({z}/\sqrt{{a}}\right)}{\left({z}/\sqrt{{a}}\right)^{\mathrm{2}} −\mathrm{1}}= \\ $$$$={z}+\frac{\sqrt{{a}}}{\mathrm{2}}\:{ln}\mid\frac{{z}−\sqrt{{a}}}{{z}+\sqrt{{a}}}\mid \\ $$$$\Rightarrow{I}=\mathrm{2}\left({J}_{\mathrm{2}} −{J}_{\mathrm{1}} \right)= \\ $$$$=\sqrt{\mathrm{2}}{ln}\mid\frac{{z}−\sqrt{\mathrm{2}}}{{z}+\sqrt{\mathrm{2}}}\mid−{ln}\mid\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\mid \\ $$$$\Rightarrow\int\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}\:\frac{{dx}}{{x}+\mathrm{3}}\:=\:\sqrt{\mathrm{2}}\:{ln}\mid\frac{\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}−\sqrt{\mathrm{2}}}{\:\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}+\sqrt{\mathrm{2}}}\mid−{ln}\mid\frac{\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}−\mathrm{1}}{\:\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}+\mathrm{1}}\mid+{C} \\ $$
Answered by Frix last updated on 20/May/25
∫((√(x+1))/((x+3)(√(x+2))))dx= =∫(dx/( (√(x+1))(√(x+2))))−2∫(dx/( (√(x+1))(√(x+2))(x+3))) (1) ∫(dx/( (√(x+1))(√(x+2))))= _(dx=2(√(x+1))(√(x+2))(dt/t))^(t=(√(x+1))+(√(x+2))) =2∫(dt/t)=2ln t =2ln ((√(x+1))+(√(x+2))) (2) −2∫(dx/( (√(x+1))(√(x+2))(x+3)))= _(dx=(√(x+1))(√(x+2))(dt/(t−3)))^(t=2(x+3+(√(x+1))(√(x+2))) ⇔ x=((t^2 −12t+20)/(4(t−3)))) =−8∫(dt/(t^2 −8))=(√2)ln ((t+2(√2))/(t−2(√2))) = =(√2)ln ∣((x+3+(√2)+(√(x+1))(√(x+2)))/(x+3−(√2)+(√(x+1))(√(x+2))))∣ ⇒ ∫((√(x+1))/((x+3)(√(x+2))))dx= =2ln ∣(√(x+1))+(√(x+2))∣ +(√2)ln ∣((x+3+(√2)+(√(x+1))(√(x+2)))/(x+3−(√2)+(√(x+1))(√(x+2))))∣ +C
$$\int\frac{\sqrt{{x}+\mathrm{1}}}{\left({x}+\mathrm{3}\right)\sqrt{{x}+\mathrm{2}}}{dx}= \\ $$$$=\int\frac{{dx}}{\:\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}}−\mathrm{2}\int\frac{{dx}}{\:\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}\left({x}+\mathrm{3}\right)} \\ $$$$\left(\mathrm{1}\right) \\ $$$$\int\frac{{dx}}{\:\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:_{{dx}=\mathrm{2}\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}\frac{{dt}}{{t}}} ^{{t}=\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{2}}} \\ $$$$=\mathrm{2}\int\frac{{dt}}{{t}}=\mathrm{2ln}\:{t}\:=\mathrm{2ln}\:\left(\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{2}}\right) \\ $$$$\left(\mathrm{2}\right) \\ $$$$−\mathrm{2}\int\frac{{dx}}{\:\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}\left({x}+\mathrm{3}\right)}= \\ $$$$\:\:\:\:\:\:\:\:\:\:_{{dx}=\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}\frac{{dt}}{{t}−\mathrm{3}}} ^{{t}=\mathrm{2}\left({x}+\mathrm{3}+\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}\right)\:\Leftrightarrow\:{x}=\frac{{t}^{\mathrm{2}} −\mathrm{12}{t}+\mathrm{20}}{\mathrm{4}\left({t}−\mathrm{3}\right)}} \\ $$$$=−\mathrm{8}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{8}}=\sqrt{\mathrm{2}}\mathrm{ln}\:\frac{{t}+\mathrm{2}\sqrt{\mathrm{2}}}{{t}−\mathrm{2}\sqrt{\mathrm{2}}}\:= \\ $$$$=\sqrt{\mathrm{2}}\mathrm{ln}\:\mid\frac{{x}+\mathrm{3}+\sqrt{\mathrm{2}}+\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}}{{x}+\mathrm{3}−\sqrt{\mathrm{2}}+\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}}\mid \\ $$$$\Rightarrow \\ $$$$\int\frac{\sqrt{{x}+\mathrm{1}}}{\left({x}+\mathrm{3}\right)\sqrt{{x}+\mathrm{2}}}{dx}= \\ $$$$=\mathrm{2ln}\:\mid\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{2}}\mid\:+\sqrt{\mathrm{2}}\mathrm{ln}\:\mid\frac{{x}+\mathrm{3}+\sqrt{\mathrm{2}}+\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}}{{x}+\mathrm{3}−\sqrt{\mathrm{2}}+\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}}\mid\:+{C} \\ $$

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