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Question Number 84123 by Roland Mbunwe last updated on 09/Mar/20

∫((5−x)/(1+(√((x−4)))))

$$\int\frac{\mathrm{5}−\boldsymbol{{x}}}{\mathrm{1}+\sqrt{\left(\boldsymbol{{x}}−\mathrm{4}\right)}} \\ $$

Commented by mathmax by abdo last updated on 09/Mar/20

I=∫ ((5−x)/(1+(√(x−4))))dx we do the changement (√(x−4))=t ⇒x−4=t^2 I =∫ ((5−(4+t^2 ))/(1+t)) (2t)dt =2 ∫ ((t(1−t^2 ))/(1+t))dt =2∫ ((t(1−t)(1+t))/(1+t))dt =2∫ (t−t^2 )dt =2((t^2 /2)−(t^3 /3))+C =t^2 −(2/3)t^3 +C =x−4 −(2/3)(x−4)^(3/2) +C

$${I}=\int\:\frac{\mathrm{5}−{x}}{\mathrm{1}+\sqrt{{x}−\mathrm{4}}}{dx}\:{we}\:{do}\:{the}\:{changement}\:\sqrt{{x}−\mathrm{4}}={t}\:\Rightarrow{x}−\mathrm{4}={t}^{\mathrm{2}} \\ $$$${I}\:=\int\:\frac{\mathrm{5}−\left(\mathrm{4}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}}\:\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int\:\:\frac{{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}}{dt} \\ $$$$=\mathrm{2}\int\:\frac{{t}\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}{dt}\:=\mathrm{2}\int\:\left({t}−{t}^{\mathrm{2}} \right){dt}\:=\mathrm{2}\left(\frac{{t}^{\mathrm{2}} }{\mathrm{2}}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\right)+{C} \\ $$$$={t}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} \:+{C}\:={x}−\mathrm{4}\:−\frac{\mathrm{2}}{\mathrm{3}}\left({x}−\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{C} \\ $$

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