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Question Number 33026 by prof Abdo imad last updated on 09/Apr/18

$$calculate{\int }_0^{\mathrm{\infty }}\frac{1+x^4}{1+x^6}dx.$$

Commented by abdo imad last updated on 09/Apr/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{1+x^4}{1+x^6}dxwehave$$$$I={\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^6}+{\int }_0^{\mathrm{\infty }}\frac{x^4}{1+x^6}dx.thech.x^6=tgive$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^6}={\int }_0^{\mathrm{\infty }}\frac{\frac{1}{6}{t}^{\frac{1}{6}-1}}{1+t}dt=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{6}-1}}{1+t}dt$$$$=\frac{1}{6}\frac{\pi }{sin\left(\frac{\pi }{6}\right)}(byusingtheresult{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}$$$$with0<a<1)$$$$=\frac{\pi }{6.\frac{1}{2}}=\frac{\pi }{3}.alsothesamech.x^6=tgivex=t^{\frac{1}{6}}$$$${\int }_0^{\mathrm{\infty }}\frac{x^4}{1+x^6}dx={\int }_0^{\mathrm{\infty }}\frac{t^{\frac{4}{6}}}{1+t}.\frac{1}{6}{t}^{\frac{1}{6}-1}dt$$$$=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{4}{6}+\frac{1}{6}-1}}{1+t}dt=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{5}{6}-1}}{1+t}dt$$$$=\frac{1}{6}\frac{\pi }{sin\left(\frac{5\pi }{6}\right)}=\frac{\pi }{6sin(\pi -\frac{\pi }{6})}=\frac{\pi }{6sin\left(\frac{\pi }{6}\right)}=\frac{\pi }{3}\Rightarrow$$$$I=\frac{\pi }{3}+\frac{\pi }{3}=\frac{2\pi }{3}.$$$$$$

Answered by Joel578 last updated on 09/Apr/18

$$I=\underset{n\to \mathrm{\infty }}{\lim }\left({\int }_0^n\frac{x^4+1}{x^6+1}dx\right)$$$$$$$$\int \frac{x^4+1}{x^6+1}dx=\int \frac{x^4+1+x^2-x^2}{(x^2+1)(x^4-x^2+1)}dx$$$$=\int \frac{1}{x^2+1}dx+\int \frac{x^2}{x^6+1}dx$$$$={\tan }^{-1}x+\int \frac{x^2}{x^6+1}dx(u=x^3\to du=3x^{2}dx)$$$$={\tan }^{-1}x+\frac{1}{3}\int \frac{1}{u^2+1}du$$$$={\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}u+C$$$$={\tan }^{-1}\left(x\right)+\frac{1}{3}{\tan }^{-1}\left(x^3\right)+C$$$$$$$$I=\underset{n\to \mathrm{\infty }}{\lim }{[{\tan }^{-1}\left(x\right)+\frac{1}{3}{\tan }^{-1}\left(x^3\right)]}_0^n$$$$=\underset{n\to \mathrm{\infty }}{\lim }({\tan }^{-1}\left(n\right)+\frac{1}{3}{\tan }^{-1}\left(n^3\right))-({\tan }^{-1}\left(0\right)+\frac{1}{3}{\tan }^{-1}\left(0\right))$$$$=\frac{\pi }{2}+\frac{1}{3}\left(\frac{\pi }{2}\right)-0$$$$=\frac{2\pi }{3}$$

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