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Question Number 126510 by benjo_mathlover last updated on 21/Dec/20

∫ (dx/((x−2)(√(x^2 +4x+8)))) ?

$$\:\int\:\frac{{dx}}{\left({x}−\mathrm{2}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}}}\:? \\ $$

Answered by liberty last updated on 21/Dec/20

Answered by Ar Brandon last updated on 21/Dec/20

I=∫(dx/((x−2)(√(x^2 +4x+8)))) Let (1/(x−2))=u ⇒ −(dx/((x−2)^2 ))=du ⇒ dx=−(du/u^2 ) x=(1/u)+2, x^2 =(1/u^2 )+(4/u)+4 I=−∫(u/( (√((1/u^2 )+(8/u)+20))))∙(du/u^2 ) =−∫(du/( (√(1+8u+20u^2 ))))=∓(1/( (√(20))))∫(du/( (√(u^2 +((2u)/5)+(1/(20)))))) =∓(1/( (√(20))))∫(du/( (√((u+(1/5))^2 +(1/(100)))))) =∓(1/( (√(20))))ln∣(u+(1/5))+(√(u^2 +((2u)/5)+(1/(20))))∣+C =∓(1/( (√(20))))ln∣((1/(x−2))+(1/5))+(√((x^2 +4x+8)/(20)))∣+C

$$\mathcal{I}=\int\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{8}}} \\ $$$$\mathrm{Let}\:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{2}}=\mathrm{u}\:\Rightarrow\:−\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{du}\:\Rightarrow\:\mathrm{dx}=−\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} } \\ $$$$\mathrm{x}=\frac{\mathrm{1}}{\mathrm{u}}+\mathrm{2},\:\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{u}}+\mathrm{4} \\ $$$$\mathcal{I}=−\int\frac{\mathrm{u}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{u}}+\mathrm{20}}}\centerdot\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} } \\ $$$$\:\:\:=−\int\frac{\mathrm{du}}{\:\sqrt{\mathrm{1}+\mathrm{8u}+\mathrm{20u}^{\mathrm{2}} }}=\mp\frac{\mathrm{1}}{\:\sqrt{\mathrm{20}}}\int\frac{\mathrm{du}}{\:\sqrt{\mathrm{u}^{\mathrm{2}} +\frac{\mathrm{2u}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{20}}}} \\ $$$$\:\:\:=\mp\frac{\mathrm{1}}{\:\sqrt{\mathrm{20}}}\int\frac{\mathrm{du}}{\:\sqrt{\left(\mathrm{u}+\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{100}}}} \\ $$$$\:\:\:=\mp\frac{\mathrm{1}}{\:\sqrt{\mathrm{20}}}\mathrm{ln}\mid\left(\mathrm{u}+\frac{\mathrm{1}}{\mathrm{5}}\right)+\sqrt{\mathrm{u}^{\mathrm{2}} +\frac{\mathrm{2u}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{20}}}\mid+{C} \\ $$$$\:\:\:=\mp\frac{\mathrm{1}}{\:\sqrt{\mathrm{20}}}\mathrm{ln}\mid\left(\frac{\mathrm{1}}{\mathrm{x}−\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{5}}\right)+\sqrt{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{8}}{\mathrm{20}}}\mid+{C} \\ $$

Answered by mathmax by abdo last updated on 21/Dec/20

I =∫ (dx/((x−2)(√(x^2 +4x+8)))) ⇒I =∫ (dx/((x−2)(√((x+2)^2 +4)))) =_(x+2=2sht) ∫ ((2ch(t))/((2sht−4)2cht))dt =(1/2)∫ (dt/(sht−2)) =(1/2)∫ (dt/(((e^t −e^(−t) )/2)−2)) =∫ (dt/(e^t −e^(−t) −4)) =_(e^t =z) ∫ (dz/(z(z−z^(−1) −4))) =∫ (dz/(z^2 −1−4z)) =∫ (dz/(z^2 −4z−1)) Δ^′ =4+1=5 ⇒z_1 =2+(√5) and z_2 =2−(√5) ⇒I=∫ (dz/((z−z_1 )(z−z_2 ))) =(1/(2(√5)))∫ ((1/(z−z_1 ))−(1/(z−z_2 )))dz =(1/(2(√5)))ln∣((z−z_1 )/(z−z_2 ))∣ +C =(1/(2(√5)))ln∣((e^t −2−(√5))/(e^t −2+(√5)))∣ +C we have t =argsh(((x+2)/2)) =ln(((x+2)/2)+(√(1+(((x+2)/2))^2 ))) ⇒e^t =(x/2)+1 +(√(1+((x/2)+1)^2 )) I =(1/(2(√5)))ln∣(((x/2)+(√(1+((x/2)+1)^2 ))−1−(√5))/((x/2)+(√(1+((x/2)+1)^2 ))−1+(√5)))∣ +C

$$\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4x}+\mathrm{8}}}\:\Rightarrow\mathrm{I}\:=\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)\sqrt{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}}} \\ $$$$=_{\mathrm{x}+\mathrm{2}=\mathrm{2sht}} \:\:\:\:\:\int\:\:\frac{\mathrm{2ch}\left(\mathrm{t}\right)}{\left(\mathrm{2sht}−\mathrm{4}\right)\mathrm{2cht}}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{dt}}{\mathrm{sht}−\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{dt}}{\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} }{\mathrm{2}}−\mathrm{2}}\:=\int\:\:\:\frac{\mathrm{dt}}{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} −\mathrm{4}}\:=_{\mathrm{e}^{\mathrm{t}} \:=\mathrm{z}} \:\:\:\int\:\:\frac{\mathrm{dz}}{\mathrm{z}\left(\mathrm{z}−\mathrm{z}^{−\mathrm{1}} −\mathrm{4}\right)} \\ $$$$=\int\:\:\frac{\mathrm{dz}}{\mathrm{z}^{\mathrm{2}} −\mathrm{1}−\mathrm{4z}}\:=\int\:\:\frac{\mathrm{dz}}{\mathrm{z}^{\mathrm{2}} −\mathrm{4z}−\mathrm{1}} \\ $$$$\Delta^{'} \:=\mathrm{4}+\mathrm{1}=\mathrm{5}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\mathrm{2}+\sqrt{\mathrm{5}}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\mathrm{2}−\sqrt{\mathrm{5}}\:\Rightarrow\mathrm{I}=\int\:\frac{\mathrm{dz}}{\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\int\:\left(\frac{\mathrm{1}}{\mathrm{z}−\mathrm{z}_{\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{z}−\mathrm{z}_{\mathrm{2}} }\right)\mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\mathrm{ln}\mid\frac{\mathrm{z}−\mathrm{z}_{\mathrm{1}} }{\mathrm{z}−\mathrm{z}_{\mathrm{2}} }\mid\:+\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\mathrm{ln}\mid\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{2}−\sqrt{\mathrm{5}}}{\mathrm{e}^{\mathrm{t}} −\mathrm{2}+\sqrt{\mathrm{5}}}\mid\:+\mathrm{C}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{t}\:=\mathrm{argsh}\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{2}}\right) \\ $$$$=\mathrm{ln}\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{2}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)\:\Rightarrow\mathrm{e}^{\mathrm{t}} \:=\frac{\mathrm{x}}{\mathrm{2}}+\mathrm{1}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{x}}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\mathrm{ln}\mid\frac{\frac{\mathrm{x}}{\mathrm{2}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{x}}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{1}−\sqrt{\mathrm{5}}}{\frac{\mathrm{x}}{\mathrm{2}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{x}}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{1}+\sqrt{\mathrm{5}}}\mid\:+\mathrm{C} \\ $$

Answered by MJS_new last updated on 21/Dec/20

∫(dx/((x−2)(√(x^2 +4x+8))))= [t=((x+2+(√(x^2 +4x+8)))/2) → dx=((2(√(x^2 +4x+8)))/(x+2+(√(x^2 +4x+8))))] =∫(dt/(t^2 −4t−1))=((√5)/(10))∫((1/(t−2−(√5)))−(1/(t−2+(√5))))dt= =((√5)/(10))ln ((t−2−(√5))/(t−2+(√5))) =((√5)/(10))ln ∣((2(2+(√5))(x+3)−(5+2(√5))(√(x^2 +4x+8)))/(x−2))∣ +C

$$\int\frac{{dx}}{\left({x}−\mathrm{2}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}}}{{x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}}}\right] \\ $$$$=\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{1}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{10}}\int\left(\frac{\mathrm{1}}{{t}−\mathrm{2}−\sqrt{\mathrm{5}}}−\frac{\mathrm{1}}{{t}−\mathrm{2}+\sqrt{\mathrm{5}}}\right){dt}= \\ $$$$=\frac{\sqrt{\mathrm{5}}}{\mathrm{10}}\mathrm{ln}\:\frac{{t}−\mathrm{2}−\sqrt{\mathrm{5}}}{{t}−\mathrm{2}+\sqrt{\mathrm{5}}}\:=\frac{\sqrt{\mathrm{5}}}{\mathrm{10}}\mathrm{ln}\:\mid\frac{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\left({x}+\mathrm{3}\right)−\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}}}{{x}−\mathrm{2}}\mid\:+{C} \\ $$

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