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Question Number 176991 by mathlove last updated on 29/Sep/22

Answered by som(math1967) last updated on 29/Sep/22

$$\int \frac{dx}{x^4+8x^2+16}$$$$=\frac{1}{8}\int \frac{8dx}{x^4+8x^2+16}$$$$=\frac{1}{8}\int \frac{\frac{8}{x^2}}{x^2+{\left(\frac{4}{x}\right)}^2+8}dx$$$$=\frac{1}{8}\int \frac{(1+\frac{4}{x^2})dx}{{(x-\frac{4}{x})}^2+{\left(4\right)}^2}-\frac{1}{8}\int \frac{(1-\frac{4}{x^2})dx}{{(x+\frac{4}{x})}^2}$$$$=\frac{1}{8}\int \frac{d(x-\frac{4}{x})}{{(x-\frac{4}{x})}^2+4^2}-\frac{1}{8}\int \frac{d(x+\frac{4}{x})}{{(x+\frac{4}{x})}^2}$$$$=\frac{1}{8}\times \frac{1}{4}{\tan }^{-1}\left(\frac{x-\frac{4}{x}}{4}\right)+\frac{1}{8}\times \frac{1}{(x+\frac{4}{x})}+C$$$$=\frac{1}{32}{\tan }^{-1}\left(\frac{x^2-4}{4x}\right)+\frac{1}{8}\times \frac{x}{(x^2+4)}+C$$

Commented by mathlove last updated on 29/Sep/22

$$thanks$$

Commented by peter frank last updated on 29/Sep/22

$$\mathrm{thank}\mathrm{you}$$

Answered by Mathspace last updated on 29/Sep/22

$$I{=}_{x=2tan\theta }\int \frac{2(1+{tan}^2\theta )d\theta }{16{(1+{tan}^2\theta )}^2}$$$$=\frac{1}{8}\int \frac{d\theta }{1+{tan}^2\theta }=\frac{1}{8}\int {cos}^2\theta d\theta$$$$=\frac{1}{16}\int (1+cos\left(2\theta \right))d\theta$$$$=\frac{1}{16}\theta +\frac{1}{32}sin\left(2\theta \right)+c$$$$=\frac{1}{16}arctan\left(\frac{x}{2}\right)+\frac{1}{32}sin\left(2arctan\left(\frac{x}{2}\right)\right)+c$$

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