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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

Consider two points, P and Q. P = (x,y) and Q=(x+δx,y+δy) Let s = length of arc from starting point δs = Length of arc PQ Since δs is very small, we can approximate arc PQ to a straight line. By Pythagoras′s theorem, (δx)^2 +(δy)^2 =(δs)^2 Dividing both sides by (δx)^2 , we get 1+(((δy)/(δx)))^2 =(((δs)/(δx)))^2 As δx→0, this gives 1+((dy/dx))^2 =((ds/dx))^2 Applying Square Root on both sides, (ds/dx)=(√(1+((dy/dx))^2 )) Therefore, we have s=∫(√(1+((dy/dx))^2 ))dx or ∫(√(1+(f′(x))^2 ))dx

$${Consider}\:{two}\:{points},\:{P}\:{and}\:{Q}. \\ $$$${P}\:=\:\left({x},{y}\right)\:{and}\:{Q}=\left({x}+\delta{x},{y}+\delta{y}\right) \\ $$$${Let}\:{s}\:=\:{length}\:{of}\:{arc}\:{from}\:{starting}\:{point} \\ $$$$\delta{s}\:=\:{Length}\:{of}\:{arc}\:{PQ} \\ $$$$ \\ $$$${Since}\:\delta{s}\:{is}\:{very}\:{small},\:{we}\:{can}\:{approximate}\:{arc}\:{PQ}\:{to}\:{a}\:{straight}\:{line}. \\ $$$${By}\:{Pythagoras}'{s}\:{theorem}, \\ $$$$\left(\delta{x}\right)^{\mathrm{2}} +\left(\delta{y}\right)^{\mathrm{2}} =\left(\delta{s}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${Dividing}\:{both}\:{sides}\:{by}\:\left(\delta{x}\right)^{\mathrm{2}} ,\:{we}\:{get} \\ $$$$\mathrm{1}+\left(\frac{\delta{y}}{\delta{x}}\right)^{\mathrm{2}} =\left(\frac{\delta{s}}{\delta{x}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${As}\:\delta{x}\rightarrow\mathrm{0},\:{this}\:{gives} \\ $$$$\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\left(\frac{{ds}}{{dx}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${Applying}\:{Square}\:{Root}\:{on}\:{both}\:{sides}, \\ $$$$\frac{{ds}}{{dx}}=\sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${Therefore},\:{we}\:{have} \\ $$$${s}=\int\sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} }{dx}\:{or}\:\int\sqrt{\mathrm{1}+\left({f}'\left({x}\right)\right)^{\mathrm{2}} }{dx} \\ $$

Answered by TonyCWX08 last updated on 27/Nov/24

∫(dx/(x^4 +5x^2 +6)) =∫(dx/((x^2 +3)(x^2 +2))) =∫(−(1/(x^2 +3))+(1/(x^2 +2)))dx =−∫((1/(x^2 +3)))dx+∫((1/(x^2 +2)))dx =−(1/( (√3)))(tan^(−1) ((x/( (√3)))))+(1/( (√2)))(tan^(−1) ((x/( (√2)))))+C

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

Answered by lepuissantcedricjunior last updated on 28/Nov/24

b)((𝛑(3(√2)−2(√3)))/(12))

$$\left.\boldsymbol{{b}}\right)\frac{\boldsymbol{\pi}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}\right)}{\mathrm{12}} \\ $$

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