Question Number 101350 by bobhans last updated on 02/Jul/20
Answered by bemath last updated on 02/Jul/20
$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right){dx}−{xy}\:{dy}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right)\:{dx}\:=\:{xy}\:{dy}\: \\ $$$$\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}}\:{dx}\:=\:\frac{{y}\:{dy}}{\mathrm{1}−{y}^{\mathrm{2}} } \\ $$$$\int\:\left({x}+\frac{\mathrm{1}}{{x}}\right)\:{dx}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{d}\left(\mathrm{1}−{y}^{\mathrm{2}} \right)}{\mathrm{1}−{y}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:+\:\mathrm{ln}\:\left({x}\right)\:+{c}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}−{y}^{\mathrm{2}} \right) \\ $$$$\Leftrightarrow{x}^{\mathrm{2}} \:+\:\mathrm{2ln}\:\left({x}\right)\:+\:\mathrm{ln}\:\left(\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)\:=\:\mathrm{C} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{ln}\left({x}^{\mathrm{2}} \left(\mathrm{1}−{y}^{\mathrm{2}} \right)\right)\:=\:{C}\: \\ $$
Commented by bobhans last updated on 02/Jul/20
$$\mathrm{yeahhh}..\bigstar\blacksquare \\ $$