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Question Number 92115 by mhmd last updated on 04/May/20

Answered by mr W last updated on 05/May/20

$$u=\frac{dy}{dx}$$$$\frac{du}{dx}+2x{(1+u)}^2=0$$$$\frac{du}{{(1+u)}^2}=-2xdx$$$$\int \frac{du}{{(1+u)}^2}=-\int 2xdx$$$$-\frac{1}{1+u}=-(x^2\pm C_1^2)$$$$u=\frac{dy}{dx}=\frac{1}{x^2\pm C_1^2}-1$$$$y=\int (\frac{1}{x^2\pm C_1^2}-1)dx$$$$y=\frac{1}{x^2\pm C_1^2}-x+C_2$$$$\Rightarrow y=\frac{1}{C_1}{\tan }^{-1}\frac{x}{C_1}-x+C_2$$$$or$$$$\Rightarrow y=\frac{1}{2C_1}\ln \mid \frac{x-C_1}{x+C_1}\mid -x+C_2$$

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