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Question Number 159226 by cortano last updated on 14/Nov/21

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\sqrt[6]{x^5}-\sqrt{x}}{(1+x^2)\ln x}dx=?$$

Answered by mindispower last updated on 14/Nov/21

$${\int }_0^{\mathrm{\infty }}\frac{x^s-x^{\frac{1}{2}}}{(1+x^2)ln\left(x\right)}.dx=f\left(s\right)$$$$f^{\prime }\left(s\right)={\int }_0^{\mathrm{\infty }}\frac{x^s}{1+x^2}dx={\int }_0^{\mathrm{\infty }}\frac{y^{\frac{s-1}{2}}}{2(1+y)}dy=\frac{\beta (\frac{1-s}{2},\frac{s+1}{2})}{2}$$$$=\frac{\pi }{2sin\left(\frac{\pi (1-s)}{2}\right)}=\frac{\pi }{2cos\left(\frac{s\pi }{2}\right)}$$$$f\left(s\right)=\underset{\frac{1}{2}}{\int }\stackrel{s}{}\frac{\pi }{2cos\left(\frac{s\pi }{2}\right)}ds$$$$={\int }_{\frac{\pi }{4}}^{\frac{s\pi }{2}}\frac{du}{cos\left(u\right)}={\int }_{\frac{\pi }{4}}^{\frac{s\pi }{2}}\frac{cos\left(u\right)du}{(1-sin\left(u\right))(1+sin\left(u\right))}$$$$=[\frac{1}{2}.ln{\left(\frac{1+sin\left(u\right)}{1-sin\left(u\right)}\right]}_{\frac{\pi }{4}}^{\frac{s\pi }{2}}]=\frac{ln\left(\frac{1+sin\left(\frac{s\pi }{2}\right)}{1-sin\left(\frac{s\pi }{2}\right)}\right)}{2}$$$$-\frac{ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)}{2}$$$$\mathrm{\Omega }=f\left(\frac{5}{6}\right)=\frac{ln\left(\frac{1+sin\left(\frac{5\pi }{12}\right)}{1-sin\left(\frac{5\pi }{12}\right)}\right)-ln\left(\frac{1+\sqrt{2}}{\sqrt{2}-1}\right)}{2}$$$$sin\left(\frac{5\pi }{12}\right)=cos\left(\frac{\pi }{12}\right)=\frac{\sqrt{2+\sqrt{3}}}{2}$$$$$$$$$$$$$$$$$$

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