intergrate 1/(-0.4v^3)
\(\int \frac{1}{-0.4v^3}\;dv\\ = \frac{1}{-0.4}\int v^{-3}\;dv\\ = \frac{-10}{4}\int v^{-3}\;dv\\ = \frac{-10}{4}\times \frac{v^{-2}}{-2}+c\\ = \frac{5}{4}\times \frac{v^{-2}}{1}+c\\ = \frac{5}{4v^2}+c\\\)
thank you
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