-VdV/(kV+g)
where k and g are constants
plz i need answer for this with illustration
and thank you all
All you have to do is some minor algebra.
VdV/(kV+g) = $\displaystyle \frac{(1/k) (kV +g)- g/k}{kV+g}$
Seperate the fraction we get.
$\displaystyle \frac{1}{k} +\frac{- g}{k(kV+g)}$
Your integral reduces to
$\displaystyle \int \frac{1}{k} +\frac{- g}{k(kV+g)} dV$
$\displaystyle \int \frac{1}{k}+\frac{- g}{k(kV+g)} dV = \frac{V}{k} - \frac{ g ln(kV +g)}{k^{2}} $