Question
Solve the following differential equation
$(1+\text{x})^2\frac{\text{dy}}{\text{dx}}-\text{x}=2\tan^{-1}\text{x}$

Answer

$(1+\text{x})^2\frac{\text{dy}}{\text{dx}}-\text{x}=2\tan^{-1}\text{x}$
$(1+\text{x})^2\frac{\text{dy}}{\text{dx}}-\text{x}=2\tan^{-1}\text{x}+\text{x}$
$\text{dy}=\Big(\frac{2\tan^{-4}\text{x}+\text{x}}{1+\text{x}^2}\Big)\text{dx}$
$\int\text{dy}=\int\Big(\frac{2\tan^{-4}\text{x}+\text{x}}{1+\text{x}^2}\Big)\text{dx}$
$\text{y}=\int(2\text{t}+\tan\text{t})\text{dt}\ [\tan^4\text{x}=\text{t}]$
$=\frac{1}{2}\log1+\text{x}^3|+(\tan^{-1}\text{x})^2+\text{C}$

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