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Question Number 66468 by mathmax by abdo last updated on 15/Aug/19

$$calculate{I}_n={\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^n+3)}^2}withn>1$$

Commented by mathmax by abdo last updated on 16/Aug/19

$$letf\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{a+x^n}witha>0\Rightarrow f\left(a\right)=\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{dx}{(1+\frac{x^n}{a})}$$$$letusethechangement\frac{x^n}{a}=u^n\Rightarrow x^n=a{u}^n\Rightarrow x=a^{\frac{1}{n}}u\Rightarrow$$$$f\left(a\right)=\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{1}{1+u^n}{a}^{\frac{1}{n}}du=a^{\frac{1}{n}-1}{\int }_0^{\mathrm{\infty }}\frac{du}{1+u^n}changementu={\alpha }^{\frac{1}{n}}$$$$give{\int }_0^{\mathrm{\infty }}\frac{1}{1+\alpha }\frac{1}{n}{\alpha }^{\frac{1}{n}-1}d\alpha =\frac{1}{n}{\int }_0^{\mathrm{\infty }}\frac{{\alpha }^{\frac{1}{n}-1}}{1+\alpha }d\alpha =\frac{1}{n}\frac{\pi }{sin\left(\frac{\pi }{n}\right)}\Rightarrow$$$$f\left(a\right)=a^{\frac{1}{n}-1}\times \frac{\pi }{nsin\left(\frac{\pi }{n}\right)}=\frac{\pi a^{\frac{1}{n}-1}}{nsin\left(\frac{\pi }{n}\right)}andwehave$$$$f^{{}^{\prime }}\left(a\right)=-{\int }_0^{\mathrm{\infty }}\frac{dx}{{(a+x^n)}^2}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dx}{{(a+x^n)}^2}=-f^{{}^{\prime }}\left(a\right)$$$$f^{{}^{\prime }}\left(a\right)=\frac{\pi (\frac{1}{n}-1)a^{\frac{1}{n}-2}}{nsin\left(\frac{\pi }{n}\right)}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dx}{{(a+x^n)}^2}=\frac{\pi (1-\frac{1}{n})a^{\frac{1}{n}-2}}{nsin\left(\frac{\pi }{n}\right)}$$$$a=3\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dx}{{(3+x^n)}^2}=\frac{\pi (1-\frac{1}{n})3^{\frac{1}{n}-2}}{nsin\left(\frac{\pi }{n}\right)}=I_n$$$$$$

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