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0-ln-x-x-x-1-2x-1-dx-




Question Number 151504 by talminator2856791 last updated on 21/Aug/21
                     ∫_0 ^( ∞)  ((ln x)/( (√x) (√(x+1)) (√(2x+1)))) dx
$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{ln}\:{x}}{\:\sqrt{{x}}\:\sqrt{{x}+\mathrm{1}}\:\sqrt{\mathrm{2}{x}+\mathrm{1}}}\:{dx} \\ $$$$\: \\ $$
Answered by Kamel last updated on 21/Aug/21
    Ω=∫_0 ^(+∞) ((Ln(x)dx)/( (√x)(√(1+x))(√(1+2x))))  Ω=^(x=(t/2)) ∫_0 ^(+∞) ((Ln(t)−Ln(2))/( (√t)(√(2+t))(√(1+t))))dt=−∫_0 ^(+∞) ((Ln(t)dt)/( (√t)(√(1+2t))(√(1+t))))−2∫_0 ^(+∞) ((Ln(2)du)/( (√(u^2 +2))(√(1+u^2 ))))  ∴ Ω=Ln(2)∫_0 ^(+∞) (du/( (√(1+u^2 ))(√(1+2u^2 ))))        =^(u=tg(t)) Ln(2)∫_0 ^(π/2) (dt/( (√(cos^2 (t)+2sin^2 (t)))))           =Ln(2)∫_0 ^(π/2) (dt/( (√(2−cos^2 (t)))))=^(θ=(π/2)−t) ((Ln(2))/( (√2)))∫_0 ^(π/2) (dθ/( (√(1−(1/2)sin^2 (θ)))))          =((K((1/2)))/( (√2)))Ln(2)
$$ \\ $$$$ \\ $$$$\Omega=\int_{\mathrm{0}} ^{+\infty} \frac{{Ln}\left({x}\right){dx}}{\:\sqrt{{x}}\sqrt{\mathrm{1}+{x}}\sqrt{\mathrm{1}+\mathrm{2}{x}}} \\ $$$$\Omega\overset{{x}=\frac{{t}}{\mathrm{2}}} {=}\int_{\mathrm{0}} ^{+\infty} \frac{{Ln}\left({t}\right)−{Ln}\left(\mathrm{2}\right)}{\:\sqrt{{t}}\sqrt{\mathrm{2}+{t}}\sqrt{\mathrm{1}+{t}}}{dt}=−\int_{\mathrm{0}} ^{+\infty} \frac{{Ln}\left({t}\right){dt}}{\:\sqrt{{t}}\sqrt{\mathrm{1}+\mathrm{2}{t}}\sqrt{\mathrm{1}+{t}}}−\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{Ln}\left(\mathrm{2}\right){du}}{\:\sqrt{{u}^{\mathrm{2}} +\mathrm{2}}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }} \\ $$$$\therefore\:\Omega={Ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{+\infty} \frac{{du}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\sqrt{\mathrm{1}+\mathrm{2}{u}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\overset{{u}={tg}\left({t}\right)} {=}{Ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dt}}{\:\sqrt{{cos}^{\mathrm{2}} \left({t}\right)+\mathrm{2}{sin}^{\mathrm{2}} \left({t}\right)}} \\ $$$$\:\:\:\:\:\:\:\:\:={Ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dt}}{\:\sqrt{\mathrm{2}−{cos}^{\mathrm{2}} \left({t}\right)}}\overset{\theta=\frac{\pi}{\mathrm{2}}−{t}} {=}\frac{{Ln}\left(\mathrm{2}\right)}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \left(\theta\right)}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{{K}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{2}}}{Ln}\left(\mathrm{2}\right) \\ $$

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