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Question Number 212052 by Spillover last updated on 28/Sep/24

Answered by Ghisom last updated on 28/Sep/24

$$\int \frac{dx}{{(x+\frac{1}{x})}^2}=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{M}\mathrm{ethod}\right]$$$$=-\frac{x}{2(x^2+1)}+\frac{1}{2}\int \frac{dx}{x^2+1}=$$$$=\frac{1}{2}\arctan x-\frac{x}{2(x^2+1)}+C$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{{(x+\frac{1}{x})}^2}=\frac{\pi }{4}$$

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