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Question Number 20872 by j.masanja06@gmail.com last updated on 05/Sep/17

if  y=[xtan^(−1) x]−[(1/2)ln(1+x^2 )]  show that (1+x^2 )y^(′′) =1

$${if}\:\:{y}=\left[{xtan}^{−\mathrm{1}} {x}\right]−\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right] \\ $$$${show}\:{that}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{''} =\mathrm{1} \\ $$

Answered by sma3l2996 last updated on 05/Sep/17

y′=tan^(−1) x+(x/(1+x^2 ))−(x/(1+x^2 ))  y′′=(1/(1+x^2 ))⇔(1+x^2 )y′′=1

$${y}'={tan}^{−\mathrm{1}} {x}+\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${y}''=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\Leftrightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}''=\mathrm{1} \\ $$

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