Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 185836 by greougoury555 last updated on 28/Jan/23

$$\left(1\right)\underset{0}{\stackrel{1}{\int }}\frac{dx}{2x^4-2x^2+1}=?$$$$\left(2\right)\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{x^{1/4}}{1+x^2}dx=?$$

Answered by Ar Brandon last updated on 28/Jan/23

$$I={\int }_0^1\frac{dx}{2x^4-2x^2+1}=\frac{1}{2}{\int }_0^1\frac{(\sqrt{2}{x}^2+1)-(\sqrt{2}{x}^2-1)}{2x^4-2x^2+1}dx$$$$=\frac{1}{2}{\int }_0^1\frac{\sqrt{2}{x}^2+1}{2x^4-2x^2+1}dx-\frac{1}{2}{\int }_0^1\frac{\sqrt{2}{x}^2-1}{2x^4-2x^2+1}dx$$$$=\frac{1}{2}{\int }_0^1\frac{\sqrt{2}+\frac{1}{x^2}}{2x^2-2+\frac{1}{x^2}}dx-\frac{1}{2}{\int }_0^1\frac{\sqrt{2}-\frac{1}{x^2}}{2x^2-2+\frac{1}{x^2}}dx$$$$=\frac{1}{2}{\int }_0^1\frac{\sqrt{2}+\frac{1}{x^2}}{{(\sqrt{2}x-\frac{1}{x})}^2+2\sqrt{2}-2}dx-\frac{1}{2}{\int }_0^1\frac{\sqrt{2}-\frac{1}{x^2}}{{(\sqrt{2}x+\frac{1}{x})}^2-2\sqrt{2}-2}dx$$$$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\sqrt{2}-1}\frac{du}{u^2+2\sqrt{2}-2}-\frac{1}{2}{\int }_{+\mathrm{\infty }}^{\sqrt{2}+1}\frac{d\mathrm{v}}{{\mathrm{v}}^2-(2\sqrt{2}+2)}$$$$=\frac{1}{2\sqrt{2\sqrt{2}-2}}{\left[\arctan \left(\frac{u}{\sqrt{2\sqrt{2}-2}}\right)\right]}_{-\mathrm{\infty }}^{\sqrt{2}-1}+\frac{1}{2\sqrt{2\sqrt{2}+2}}{\left[\mathrm{argcoth}\left(\frac{\mathrm{v}}{\sqrt{2\sqrt{2}+2}}\right)\right]}_{+\mathrm{\infty }}^{\sqrt{2}+1}$$$$=\frac{1}{2\sqrt{2(\sqrt{2}-1)}}[\arctan \left(\sqrt{\frac{\sqrt{2}-1}{2}}\right)-(-\frac{\pi }{2})]+\frac{1}{4\sqrt{2\sqrt{2}+2}}{[\ln \mid \frac{\mathrm{v}+\sqrt{2\sqrt{2}+2}}{\mathrm{v}-\sqrt{2\sqrt{2}+2}}\mid ]}_{+\mathrm{\infty }}^{\sqrt{2}+1}$$$$=\frac{1}{2\sqrt{2(\sqrt{2}-1)}}[\arctan \left(\sqrt{\frac{\sqrt{2}-1}{2}}\right)+\frac{\pi }{2}]+\frac{1}{4\sqrt{2\sqrt{2}+2}}\ln \mid \frac{\sqrt{2}+1+\sqrt{2\sqrt{2}+2}}{\sqrt{2}+1-\sqrt{2\sqrt{2}+2}}\mid$$

Answered by Ar Brandon last updated on 28/Jan/23

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{x^{\frac{1}{4}}}{1+x^2}dx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^{-\frac{3}{8}}}{1+t}dt=\frac{1}{2}\beta (\frac{5}{8},\frac{3}{8})$$$$=\frac{1}{2}\cdot \frac{\pi }{\sin \left(\frac{3\pi }{8}\right)}=\frac{\pi }{2}\mathrm{cosec}\left(\frac{3\pi }{8}\right)$$

Answered by cortano1 last updated on 29/Jan/23

$$\left(2\right)\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\sqrt[4]{x}}{1+x^2}dx=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{4x^4}{1+x^8}dx$$$$=\underset{\mathrm{\infty }}{\stackrel{0}{\int }}\frac{4x^{-4}}{x^{-8}+1}(-x^{-2}dx)$$$$=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{4x^2}{1+x^8}dx$$$$2I=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{4x^4+4x^2}{1+x^8}dx$$$$I=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{2(x^4+x^2)}{1+x^8}dx=2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{1+x^{-2}}{x^4+x^{-4}}dx$$$$I=2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{d(x-x^{-1})}{{(x^2+x^{-2})}^2-2}$$$$I=2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dt}{t^4+4t^2+2};t=x-x^{-1}$$$$I=\frac{\sqrt{2}}{\sqrt{\sqrt{2}(\sqrt{2+1)}}}{\left[\arctan \left(\frac{w}{\sqrt{4+\sqrt{2\sqrt{2}}}}\right)\right]}_{-\mathrm{\infty }}^{\mathrm{\infty }},w=t-\sqrt{2}{t}^{-1}$$$$I=\frac{\pi }{\sqrt{\sqrt{2}(\sqrt{2}+1)}}$$

Answered by Mathspace last updated on 30/Jan/23

$$I={\int }_0^{\mathrm{\infty }}\frac{x^{\frac{1}{4}}}{1+x^2}dx(x=t^{\frac{1}{2}})$$$$={\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{8}}}{1+t}.\frac{1}{2}{t}^{-\frac{1}{2}}dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^{-\frac{3}{8}}}{1+t}dt=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^{-\frac{3}{8}-1+1}}{1+t}dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{5}{8}-1}}{1+t}dt=\frac{1}{2}.\frac{\pi }{sin\left(\frac{5\pi }{8}\right)}$$$$=\frac{\pi }{2sin(\frac{\pi }{2}+\frac{\pi }{8})}=\frac{\pi }{2cos\left(\frac{\pi }{8}\right)}$$$$=\frac{\pi }{2.\frac{\sqrt{2+\sqrt{2}}}{2}}=\frac{\pi }{\sqrt{2+\sqrt{2}}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com