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Question Number 30390 by mondodotto@gmail.com last updated on 21/Feb/18

Commented by abdo imad last updated on 21/Feb/18

$$wehave{x}^2+4x+3={(x+2)}^2-1tbech.x+2=chtgive$$$$I=\int \sqrt{{ch}^{2}t-1}shtdt=\int {sh}^{2}tdt=\int \frac{1-ch\left(2t\right)}{2}dt$$$$=\frac{1}{2}t-\frac{1}{2}\int ch\left(2t\right)dt=\frac{t}{2}-\frac{1}{4}sh\left(2t\right)$$$$=\frac{t}{2}-\frac{1}{2}sh\left(t\right)ch\left(t\right)=\frac{1}{2}argch(x+2)-\frac{1}{2}\sqrt{{(x+2)}^2-1}(x+2)$$$$I=\frac{1}{2}ln({(x+2)}^2+\sqrt{{(x+2)}^2-1})-\frac{1}{2}(x+2)\sqrt{{(x+2)}^2-1}+\lambda$$

Commented by mondodotto@gmail.com last updated on 22/Feb/18

$$\mathrm{thAnx}$$

Commented by mondodotto@gmail.com last updated on 22/Feb/18

$$\mathrm{what}\mathrm{does}\mathrm{it}\mathrm{mean}?\mathrm{ch}\mathrm{and}\mathrm{sh}????$$

Commented by prof Abdo imad last updated on 22/Feb/18

$$ch\left(x\right)=\frac{e^x+e^{-x}}{2},sh\left(x\right)=\frac{e^x-e^{-x}}{2}$$$$th\left(x\right)=\frac{shx}{chx}withxfromRandthesame$$$$definitiononset\mathbb{C}.$$

Commented by mondodotto@gmail.com last updated on 22/Feb/18

$$\mathrm{now}\mathrm{i}\mathrm{undstnd}!$$

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