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The integral can be written as: $$\frac{1}{k} \int \frac{dv}{a^2+v^2}=-\int dx$$ Where $a=\sqrt\frac{\mu g}{k}$ Now, this is the standard $\arctan$ integral, which evaluates to: $$\frac{1}{ka} \tan^{-1} {v\over a}=-x+C$$ Now you can put $a$ back.
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