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Question Number 127958 by mnjuly1970 last updated on 03/Jan/21

$$...nicecalculus...$$$$calculate$$$$\mathrm{\Omega }={\int }_1^{\mathrm{\infty }}\frac{ln(x^4-2x^2+2)}{x\sqrt{x^2-1}}dx=?$$$$$$

Answered by Dwaipayan Shikari last updated on 03/Jan/21

$${\int }_1^{\mathrm{\infty }}\frac{log(x^4-2x^2+2)}{x\sqrt{x^2-1}}dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{log({sec}^4\theta -2{sec}^2\theta +2)}{sec\theta tan\theta }sec\theta tan\theta$$$$={\int }_0^{\frac{\pi }{2}}log(1-2{cos}^2\theta +2{cos}^4\theta )-4{\int }_0^{\frac{\pi }{2}}log\left(cos\theta \right)d\theta$$$$={\int }_0^{\frac{\pi }{2}}log(1-\frac{1}{2}{sin}^{2}2\theta )+2\pi log\left(2\right)$$$$={\int }_0^{\frac{\pi }{2}}log({cos}^{2}2\theta +\frac{1}{2}{sin}^{2}2\theta )+2\pi log\left(2\right)$$$$=\pi log\left(\frac{1+\frac{1}{\sqrt{2}}}{2}\right)+\pi log\left(4\right)=\pi log(2+\sqrt{2})$$$$Lemma$$$${\int }_0^{\frac{\pi }{2}}log(a^2{cos}^2\theta +b^2{sin}^2\theta )=\pi log\left(\frac{a+b}{2}\right)$$

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