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Question Number 66467 by mathmax by abdo last updated on 15/Aug/19

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

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

$$wehaveproved{\int }_0^{\mathrm{\infty }}\frac{dx}{{(a+x^n)}^2}=\frac{\pi }{nsin\left(\frac{\pi }{n}\right)}(1-\frac{1}{n})a^{\frac{1}{n}-2}$$ $$a=n\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dx}{{(n+x^n)}^n}=\frac{\pi }{nsin\left(\frac{\pi }{n}\right)}(1-\frac{1}{n})n^{\frac{1}{n}-2}$$ $$=\frac{\pi (n-1)}{n^{2}sin\left(\frac{\pi }{n}\right)}\left({}^n\sqrt{n}\right)\times \frac{1}{n^2}=\frac{\pi (n-1)\left({}^n\sqrt{n}\right)}{n^{4}sin\left(\frac{\pi }{n}\right)}withn>1$$

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