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Question Number 214072 by Mwamba last updated on 26/Nov/24

Answered by TonyCWX08 last updated on 27/Nov/24

Answered by TonyCWX08 last updated on 27/Nov/24

$$Considertwopoints,PandQ.$$$$P=(x,y)andQ=(x+\delta x,y+\delta y)$$$$Lets=lengthofarcfromstartingpoint$$$$\delta s=LengthofarcPQ$$$$$$$$Since\delta sisverysmall,wecanapproximatearcPQtoastraightline.$$$$By{Pythagoras}^{\prime }stheorem,$$$${\left(\delta x\right)}^2+{\left(\delta y\right)}^2={\left(\delta s\right)}^2$$$$$$$$Dividingbothsidesby{\left(\delta x\right)}^2,weget$$$$1+{\left(\frac{\delta y}{\delta x}\right)}^2={\left(\frac{\delta s}{\delta x}\right)}^2$$$$$$$$As\delta x\to 0,thisgives$$$$1+{\left(\frac{dy}{dx}\right)}^2={\left(\frac{ds}{dx}\right)}^2$$$$$$$$ApplyingSquareRootonbothsides,$$$$\frac{ds}{dx}=\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$$$$$$$$Therefore,wehave$$$$s=\int \sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dxor\int \sqrt{1+{\left(f^{\prime }\left(x\right)\right)}^2}dx$$

Answered by TonyCWX08 last updated on 27/Nov/24

$$\int \frac{dx}{x^4+5x^2+6}$$$$=\int \frac{dx}{(x^2+3)(x^2+2)}$$$$=\int (-\frac{1}{x^2+3}+\frac{1}{x^2+2})dx$$$$=-\int \left(\frac{1}{x^2+3}\right)dx+\int \left(\frac{1}{x^2+2}\right)dx$$$$=-\frac{1}{\sqrt{3}}\left({\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)\right)+\frac{1}{\sqrt{2}}\left({\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)\right)+C$$

Answered by lepuissantcedricjunior last updated on 28/Nov/24

$$\mathit{b})\frac{\mathit{\pi }(3\sqrt{2}-2\sqrt{3})}{12}$$

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