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Question Number 30769 by abdo imad last updated on 25/Feb/18
find the value of I= ∫_0 ^1      (dx/((x+1)^2 (√(x^2  +2x +2)))) .
$${find}\:{the}\:{value}\:{of}\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}}}\:. \\ $$
Commented by abdo imad last updated on 27/Feb/18
we have x^2  +2x+2=(x+1)^2  +1 the ch=x+1=t give  I= ∫_1 ^2     (dt/(t^2 (√(1+t^2 )))) the we can use the ch.t=tanθ ⇒  I= ∫_(π/4) ^(artan2)  ((1+tan^2 θ)/(tan^2 θ(√(1+tan^2 θ))))dθ = ∫_(π/4) ^(arctan2)    ((cos^2 θ)/(sin^2 θ)) (1/(cosθ))dθ  = ∫_(π/4) ^(arctan2)     ((cosθ)/(sin^2 θ)) dθ  =[−(1/(sinθ))]_(π/4) ^(arctan2)   =(√2) − (1/(sin(arctan2))) but we have the formula  sin(arctanx)= (x/( (√(1+x^2 ))))( for proof put x=tanθ)⇒  sin(arctan2)= (2/( (√5)))  ⇒ I=(√2)  −(2/( (√5))) .
$${we}\:{have}\:{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}=\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\:{the}\:{ch}={x}+\mathrm{1}={t}\:{give} \\ $$$${I}=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:{the}\:{we}\:{can}\:{use}\:{the}\:{ch}.{t}={tan}\theta\:\Rightarrow \\ $$$${I}=\:\int_{\frac{\pi}{\mathrm{4}}} ^{{artan}\mathrm{2}} \:\frac{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{{tan}^{\mathrm{2}} \theta\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}}{d}\theta\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{{arctan}\mathrm{2}} \:\:\:\frac{{cos}^{\mathrm{2}} \theta}{{sin}^{\mathrm{2}} \theta}\:\frac{\mathrm{1}}{{cos}\theta}{d}\theta \\ $$$$=\:\int_{\frac{\pi}{\mathrm{4}}} ^{{arctan}\mathrm{2}} \:\:\:\:\frac{{cos}\theta}{{sin}^{\mathrm{2}} \theta}\:{d}\theta\:\:=\left[−\frac{\mathrm{1}}{{sin}\theta}\right]_{\frac{\pi}{\mathrm{4}}} ^{{arctan}\mathrm{2}} \\ $$$$=\sqrt{\mathrm{2}}\:−\:\frac{\mathrm{1}}{{sin}\left({arctan}\mathrm{2}\right)}\:{but}\:{we}\:{have}\:{the}\:{formula} \\ $$$${sin}\left({arctanx}\right)=\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\left(\:{for}\:{proof}\:{put}\:{x}={tan}\theta\right)\Rightarrow \\ $$$${sin}\left({arctan}\mathrm{2}\right)=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\:\Rightarrow\:{I}=\sqrt{\mathrm{2}}\:\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:. \\ $$

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