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 | ||
$${Consider}\:{two}\:{points},\:{P}\:{and}\:{Q}. \\ $$$${P}\:=\:\left({x},{y}\right)\:{and}\:{Q}=\left({x}+\delta{x},{y}+\delta{y}\right) \\ $$$${Let}\:{s}\:=\:{length}\:{of}\:{arc}\:{from}\:{starting}\:{point} \\ $$$$\delta{s}\:=\:{Length}\:{of}\:{arc}\:{PQ} \\ $$$$ \\ $$$${Since}\:\delta{s}\:{is}\:{very}\:{small},\:{we}\:{can}\:{approximate}\:{arc}\:{PQ}\:{to}\:{a}\:{straight}\:{line}. \\ $$$${By}\:{Pythagoras}'{s}\:{theorem}, \\ $$$$\left(\delta{x}\right)^{\mathrm{2}} +\left(\delta{y}\right)^{\mathrm{2}} =\left(\delta{s}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${Dividing}\:{both}\:{sides}\:{by}\:\left(\delta{x}\right)^{\mathrm{2}} ,\:{we}\:{get} \\ $$$$\mathrm{1}+\left(\frac{\delta{y}}{\delta{x}}\right)^{\mathrm{2}} =\left(\frac{\delta{s}}{\delta{x}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${As}\:\delta{x}\rightarrow\mathrm{0},\:{this}\:{gives} \\ $$$$\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\left(\frac{{ds}}{{dx}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${Applying}\:{Square}\:{Root}\:{on}\:{both}\:{sides}, \\ $$$$\frac{{ds}}{{dx}}=\sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${Therefore},\:{we}\:{have} \\ $$$${s}=\int\sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} }{dx}\:{or}\:\int\sqrt{\mathrm{1}+\left({f}'\left({x}\right)\right)^{\mathrm{2}} }{dx} \\ $$ | ||
Answered by TonyCWX08 last updated on 27/Nov/24 | ||
$$\int\frac{{dx}}{{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{6}} \\ $$$$=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)} \\ $$$$=\int\left(−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{3}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}}\right){dx} \\ $$$$=−\int\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{3}}\right){dx}+\int\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}}\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{\:\sqrt{\mathrm{3}}}\right)\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{\:\sqrt{\mathrm{2}}}\right)\right)+{C} \\ $$ | ||
Answered by lepuissantcedricjunior last updated on 28/Nov/24 | ||
$$\left.\boldsymbol{{b}}\right)\frac{\boldsymbol{\pi}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}\right)}{\mathrm{12}} \\ $$ | ||