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Question Number 82440 by mathmax by abdo last updated on 21/Feb/20
find ∫ ((x+1)/(x+2))(√((1−x)/(1+x)))dx
$${find}\:\int\:\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{dx} \\ $$
Commented by abdomathmax last updated on 24/Feb/20
let A =∫((x+1)/(x+2))(√((1−x)/(1+x)))dx changement (√((1−x)/(1+x)))=t hive ((1−x)/(1+x))=t^2 ⇒1−x=t^2 +xt^2 ⇒1−t^2 =x(1+t^2 ) ⇒ x=((1−t^2 )/(1+t^2 )) ⇒dx =((−2t(1+t^2 )−(1−t^2 )2t)/((1+t^2 )^2 ))dt =((−2t−2t^3 −2t+2t^3 )/((1+t^2 )^2 ))dt=((−4t)/((1+t^2 )^2 ))dt x+1 =((1−t^2 )/(1+t^2 ))+1 =((1−t^2 +1+t^2 )/(1+t^2 )) =(2/(1+t^2 )) x+2 =((1−t^2 )/(1+t^2 )) +2 =((1−t^2 +2+2t^2 )/(1+t^2 )) =((3+t^2 )/(1+t^2 )) ⇒ I =∫ (2/(1+t^2 ))×((1+t^2 )/(3+t^2 ))×t ×((−4t)/((1+t^2 )^2 ))dt =∫ ((−8t^2 )/((t^2 +3)(t^2 +1)^2 ))dt let decpmpose F(t)=((−8t^2 )/((t^2 +3)(t^2 +1)^2 )) ⇒ F(t) =((at+b)/(t^2 +3)) +((ct+d)/(t^2 +1)) +((dt +f)/((t^2 +1)^2 )) ....be vontinued...
$${let}\:{A}\:=\int\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{dx}\:\:{changement}\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}={t} \\ $$$${hive}\:\:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}={t}^{\mathrm{2}} \:\Rightarrow\mathrm{1}−{x}={t}^{\mathrm{2}} \:+{xt}^{\mathrm{2}} \:\Rightarrow\mathrm{1}−{t}^{\mathrm{2}} ={x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${x}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow{dx}\:=\frac{−\mathrm{2}{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\mathrm{2}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$=\frac{−\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{3}} −\mathrm{2}{t}+\mathrm{2}{t}^{\mathrm{3}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}=\frac{−\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$${x}+\mathrm{1}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{1}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} \:+\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${x}+\mathrm{2}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\mathrm{2}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} \:+\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:=\frac{\mathrm{3}+{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{I}\:=\int\:\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }×\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{3}+{t}^{\mathrm{2}} }×{t}\:×\frac{−\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$=\int\:\:\frac{−\mathrm{8}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\:{let}\:{decpmpose} \\ $$$${F}\left({t}\right)=\frac{−\mathrm{8}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{{at}+{b}}{{t}^{\mathrm{2}} \:+\mathrm{3}}\:+\frac{{ct}+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{dt}\:+{f}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$….{be}\:{vontinued}… \\ $$

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