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Question Number 193768 by cortano12 last updated on 19/Jun/23

$$\mathrm{f}\left(\mathrm{x}\right)=2+\underset{0}{\stackrel{\mathrm{x}}{\int }}{(2\mathrm{t}+\mathrm{f}\left(\mathrm{t}\right))}^2\mathrm{dt}$$$$\mathrm{then}\underset{-1}{\stackrel{2}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=$$

Answered by gatocomcirrose last updated on 20/Jun/23

$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={(2\mathrm{x}+\mathrm{f}\left(\mathrm{x}\right))}^2$$$$\mathrm{v}\left(\mathrm{x}\right)=2\mathrm{x}+\mathrm{y}\left(\mathrm{x}\right)\Rightarrow {\mathrm{v}}^{\prime }\left(\mathrm{x}\right)=2+{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=2+\mathrm{v}{\left(\mathrm{x}\right)}^2$$$$\frac{\mathrm{dv}}{2+{\mathrm{v}}^2}=\mathrm{dx}\Rightarrow \frac{1}{\sqrt{2}}\arctan \left(\frac{\mathrm{v}\left(\mathrm{x}\right)}{\sqrt{2}}\right)=\mathrm{x}+{\mathrm{c}}_1$$$$\Rightarrow \mathrm{v}\left(\mathrm{x}\right)=\sqrt{2}\tan (\sqrt{2}\mathrm{x}+\mathrm{C})$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\sqrt{2}\tan (\sqrt{2}\mathrm{x}+\mathrm{C})-2\mathrm{x}$$$$$$$${\int }_{-1}^2(\sqrt{2}\tan (\sqrt{2}\mathrm{x}+\mathrm{C})-2\mathrm{x})\mathrm{dx}$$$$=\sqrt{2}{\int }_{-1}^2\tan (\sqrt{2}\mathrm{x}+\mathrm{C})\mathrm{dx}-3$$$$={[-\ln \mid \cos (\sqrt{2}\mathrm{x}+\mathrm{C})\mid ]}_{-1}^2-3$$$$=\ln \mid \cos (\mathrm{C}-\sqrt{2})\mid -\ln \mid \cos (2\sqrt{2}+\mathrm{C})\mid -3$$$$=\ln \mid \frac{\cos (\mathrm{C}-\sqrt{2})}{\cos (\mathrm{C}+2\sqrt{2})}\mid -3,\mathrm{C}\in \mathbb{R}$$

Commented by mr W last updated on 20/Jun/23

$$fromf\left(x\right){\mid }_{x=0}=2youcangetC.$$

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