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Question Number 151504 by talminator2856791 last updated on 21/Aug/21

$$$$$${\int }_0^{\mathrm{\infty }}\frac{\ln x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}dx$$$$$$

Answered by Kamel last updated on 21/Aug/21

$$$$$$$$$$\mathrm{\Omega }={\int }_0^{+\mathrm{\infty }}\frac{Ln\left(x\right)dx}{\sqrt{x}\sqrt{1+x}\sqrt{1+2x}}$$$$\mathrm{\Omega }\stackrel{x=\frac{t}{2}}{=}{\int }_0^{+\mathrm{\infty }}\frac{Ln\left(t\right)-Ln\left(2\right)}{\sqrt{t}\sqrt{2+t}\sqrt{1+t}}dt=-{\int }_0^{+\mathrm{\infty }}\frac{Ln\left(t\right)dt}{\sqrt{t}\sqrt{1+2t}\sqrt{1+t}}-2{\int }_0^{+\mathrm{\infty }}\frac{Ln\left(2\right)du}{\sqrt{u^2+2}\sqrt{1+u^2}}$$$$\therefore \mathrm{\Omega }=Ln\left(2\right){\int }_0^{+\mathrm{\infty }}\frac{du}{\sqrt{1+u^2}\sqrt{1+2u^2}}$$$$\stackrel{u=tg\left(t\right)}{=}Ln\left(2\right){\int }_0^{\frac{\pi }{2}}\frac{dt}{\sqrt{{cos}^2\left(t\right)+2{sin}^2\left(t\right)}}$$$$=Ln\left(2\right){\int }_0^{\frac{\pi }{2}}\frac{dt}{\sqrt{2-{cos}^2\left(t\right)}}\stackrel{\theta =\frac{\pi }{2}-t}{=}\frac{Ln\left(2\right)}{\sqrt{2}}{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{1-\frac{1}{2}{sin}^2\left(\theta \right)}}$$$$=\frac{K\left(\frac{1}{2}\right)}{\sqrt{2}}Ln\left(2\right)$$

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