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Question Number 91421 by  M±th+et+s last updated on 30/Apr/20

∫_1 ^3 (1/(x(√(3x^2 +2x−1))))dx

$$\int_{\mathrm{1}} ^{\mathrm{3}} \frac{\mathrm{1}}{{x}\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}{dx} \\ $$

Commented by abdomathmax last updated on 30/Apr/20

I =∫_1 ^3 (dx/(x(√(3x^2 +2x−1)))) 3x^2 +2x−1=0→ Δ^′ =1+3 =4 ⇒x_1 =((−1+2)/3)=(1/3) x_2 =((−1−2)/3) =−1 ⇒I =∫_1 ^3 (dx/(x(√(3(x−(1/3))(x+1))))) = ∫_1 ^3 (dx/(x(√((3x−1)(x+1))))) =∫_1 ^3 (dx/(x(√(x+1))(√(3x−1)))) we do the changement (√(x+1))=t ⇒x+1 =t^2 ⇒ I =∫_(√2) ^2 ((2tdt)/((t^2 −1)(√(3(t^2 −1)−1)))) =∫_(√2) ^2 ((2t dt)/((t^2 −1)(√(3t^2 −4)))) =(2/(√3))∫_(√2) ^2 ((tdt)/((t^2 −1)(√(t^2 −(4/3))))) =_(t =(2/(√3)) ch(u)→u=argch(((t(√3))/2))) (2/(√3))∫_(argch(((√6)/2))) ^(argch((√3))) ((2/(√3)))^2 ×((chu shu)/(((4/(√3))ch^2 u−1)(2/(√3)) shu))du =(4/(√3)) ∫_(argch(((√6)/2))) ^(argch((√3))) ((shu)/((4ch^2 u−(√3))))du =(4/(√3)) ∫_(argch(((√6)/2))) ^(argch((√3))) ((shu)/((4×((1+ch(2u))/2)−(√3))))du =(4/(√3)) ∫_(argch(((√6)/2))) ^(argch((√3))) ((shu)/((2ch(2u)+2−(√3))))du =(4/(√3)) ∫_(argch(((√6)/2))) ^(argch((√3))) (((e^u −e^(−u) )/2)/(e^(2u) +e^(−2u) +2−(√3)))du =(2/(√3)) ∫_(((√6)/2)+(√((6/4)−1))) ^((√3)+(√2)) ((z−z^(−1) )/(z^2 +z^(−2) +2−(√3))) (dz/z) (z=e^u ) =(2/(√3)) ∫_(((√6)/2)+((√2)/2)) ^((√3)+(√2)) ((z^2 −1)/(z^4 +1+(2−(√3))z^2 ))dz =(2/(√3))∫_(((√6)/2)+((√2)/2)) ^((√3)+(√2)) ((z^2 −1)/(z^4 +(2−(√3))z^2 +1))dx let ddcompose F(z) =((z^2 −1)/(z^4 +(2−(√3))z^2 +1)) ...be continued...

$${I}\:=\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\frac{{dx}}{{x}\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{1}=\mathrm{0}\rightarrow\:\Delta^{'} =\mathrm{1}+\mathrm{3}\:=\mathrm{4}\:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{2}}{\mathrm{3}}\:=−\mathrm{1}\:\Rightarrow{I}\:=\int_{\mathrm{1}} ^{\mathrm{3}} \:\frac{{dx}}{{x}\sqrt{\mathrm{3}\left({x}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left({x}+\mathrm{1}\right)}} \\ $$$$=\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\frac{{dx}}{{x}\sqrt{\left(\mathrm{3}{x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)}}\:=\int_{\mathrm{1}} ^{\mathrm{3}} \:\frac{{dx}}{{x}\sqrt{{x}+\mathrm{1}}\sqrt{\mathrm{3}{x}−\mathrm{1}}} \\ $$$${we}\:{do}\:{the}\:{changement}\:\sqrt{{x}+\mathrm{1}}={t}\:\Rightarrow{x}+\mathrm{1}\:={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \:\:\frac{\mathrm{2}{tdt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{3}\left({t}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{1}}} \\ $$$$=\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \:\:\frac{\mathrm{2}{t}\:{dt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}}\:=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \:\frac{{tdt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}}} \\ $$$$=_{{t}\:=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:{ch}\left({u}\right)\rightarrow{u}={argch}\left(\frac{{t}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \:\:\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\int_{{argch}\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)} ^{{argch}\left(\sqrt{\mathrm{3}}\right)} \:\left(\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} ×\frac{{chu}\:{shu}}{\left(\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}{ch}^{\mathrm{2}} {u}−\mathrm{1}\right)\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:{shu}}{du} \\ $$$$=\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}\:\int_{{argch}\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)} ^{{argch}\left(\sqrt{\mathrm{3}}\right)} \:\frac{{shu}}{\left(\mathrm{4}{ch}^{\mathrm{2}} {u}−\sqrt{\mathrm{3}}\right)}{du} \\ $$$$=\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}\:\int_{{argch}\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)} ^{{argch}\left(\sqrt{\mathrm{3}}\right)} \:\:\frac{{shu}}{\left(\mathrm{4}×\frac{\mathrm{1}+{ch}\left(\mathrm{2}{u}\right)}{\mathrm{2}}−\sqrt{\mathrm{3}}\right)}{du} \\ $$$$=\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}\:\int_{{argch}\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)} ^{{argch}\left(\sqrt{\mathrm{3}}\right)} \:\:\:\frac{{shu}}{\left(\mathrm{2}{ch}\left(\mathrm{2}{u}\right)+\mathrm{2}−\sqrt{\mathrm{3}}\right)}{du} \\ $$$$=\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}\:\int_{{argch}\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)} ^{{argch}\left(\sqrt{\mathrm{3}}\right)} \:\:\frac{\frac{{e}^{{u}} −{e}^{−{u}} }{\mathrm{2}}}{{e}^{\mathrm{2}{u}} \:+{e}^{−\mathrm{2}{u}} \:+\mathrm{2}−\sqrt{\mathrm{3}}}{du} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\int_{\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}+\sqrt{\frac{\mathrm{6}}{\mathrm{4}}−\mathrm{1}}} ^{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}} \:\:\:\:\frac{{z}−{z}^{−\mathrm{1}} }{{z}^{\mathrm{2}} \:+{z}^{−\mathrm{2}} \:+\mathrm{2}−\sqrt{\mathrm{3}}}\:\frac{{dz}}{{z}}\:\:\:\left({z}={e}^{{u}} \right) \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\int_{\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} ^{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}} \:\:\:\:\:\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}^{\mathrm{4}} \:+\mathrm{1}+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){z}^{\mathrm{2}} }{dz} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\int_{\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} ^{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}} \:\:\:\:\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}^{\mathrm{4}} \:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){z}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:{let}\:{ddcompose} \\ $$$${F}\left({z}\right)\:=\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}^{\mathrm{4}} \:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){z}^{\mathrm{2}} +\mathrm{1}}\:...{be}\:{continued}... \\ $$$$ \\ $$$$ \\ $$

Commented by jagoll last updated on 30/Apr/20

∫ (dx/(x (√(x^2 (3+(2/x)−(1/x^2 )))))) ∫ (dx/(x^2 (√(−((1/x^2 )−(2/x)+1)+4)))) ∫ (dx/(x^2 (√(4−((1/x)−1)^2 )))) [ let (1/x)−1 = t (dx/x^2 ) = −dt ∫ ((−dt)/(√(4−t^2 ))) . now easy to solve

$$\int\:\frac{{dx}}{{x}\:\sqrt{{x}^{\mathrm{2}} \left(\mathrm{3}+\frac{\mathrm{2}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}} \\ $$$$\int\:\frac{{dx}}{{x}^{\mathrm{2}} \:\sqrt{−\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{2}}{{x}}+\mathrm{1}\right)+\mathrm{4}}} \\ $$$$\int\:\frac{{dx}}{{x}^{\mathrm{2}} \:\sqrt{\mathrm{4}−\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$\left[\:{let}\:\frac{\mathrm{1}}{{x}}−\mathrm{1}\:=\:{t}\:\right. \\ $$$$\frac{{dx}}{{x}^{\mathrm{2}} }\:=\:−{dt}\: \\ $$$$\int\:\frac{−{dt}}{\sqrt{\mathrm{4}−{t}^{\mathrm{2}} }}\:.\:{now}\:{easy}\:{to}\:{solve} \\ $$

Answered by jagoll last updated on 30/Apr/20

Commented by  M±th+et+s last updated on 30/Apr/20

nice work thanks

$${nice}\:{work}\:{thanks} \\ $$

Answered by MJS last updated on 01/May/20

x^2 (3x^2 +2x−1)=x^2 (x+1)(3x−1) let x=(1/(at+b)) −(((at+b−3)(at+b+1))/((at+b)^4 )) at+b−3=t−c ⇔ (a−1)t+b+c−3=0 at+b+1=t+c ⇔ (a−1)t+b−c+1=0 ⇒ a=1 b+c−3=0∧b−c+1=0 ⇒ b=1∧k=2 −(((t−2)(t+2))/((t+1)^4 )) let t=2u −((4(u−1)(u+1))/((2u+1)^4 )) ⇒ let x=(1/(2sin u+1)) ⇔ u=arcsin ((x−1)/(2x)) → dx=x(√(3x^2 +2x−1))du ∫(dx/(x(√(3x^2 +2x−1))))=∫du=u=arcsin ((x−1)/(2x)) +C

$${x}^{\mathrm{2}} \left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}\right)={x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)\left(\mathrm{3}{x}−\mathrm{1}\right) \\ $$$$\mathrm{let}\:{x}=\frac{\mathrm{1}}{{at}+{b}} \\ $$$$−\frac{\left({at}+{b}−\mathrm{3}\right)\left({at}+{b}+\mathrm{1}\right)}{\left({at}+{b}\right)^{\mathrm{4}} } \\ $$$${at}+{b}−\mathrm{3}={t}−{c}\:\Leftrightarrow\:\left({a}−\mathrm{1}\right){t}+{b}+{c}−\mathrm{3}=\mathrm{0} \\ $$$${at}+{b}+\mathrm{1}={t}+{c}\:\Leftrightarrow\:\left({a}−\mathrm{1}\right){t}+{b}−{c}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:{a}=\mathrm{1} \\ $$$${b}+{c}−\mathrm{3}=\mathrm{0}\wedge{b}−{c}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:{b}=\mathrm{1}\wedge{k}=\mathrm{2} \\ $$$$−\frac{\left({t}−\mathrm{2}\right)\left({t}+\mathrm{2}\right)}{\left({t}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\mathrm{let}\:{t}=\mathrm{2}{u} \\ $$$$−\frac{\mathrm{4}\left({u}−\mathrm{1}\right)\left({u}+\mathrm{1}\right)}{\left(\mathrm{2}{u}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\Rightarrow \\ $$$$\mathrm{let}\:{x}=\frac{\mathrm{1}}{\mathrm{2sin}\:{u}+\mathrm{1}}\:\Leftrightarrow\:{u}=\mathrm{arcsin}\:\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\:\rightarrow\:{dx}={x}\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}{du} \\ $$$$\int\frac{{dx}}{{x}\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}=\int{du}={u}=\mathrm{arcsin}\:\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\:+{C} \\ $$

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