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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.The DV2T Viscometer features a 5-inch color display to guide users through test creation and data gathering for fast and easy viscosity measurements. The DV2T also offers powerful programming capabilities and results analysis including data averaging and QC limits with alarms.