\((dL)/(L-a)= k(x+b)dx\)

\((dL)/(L-a)= (kx+bk)dx\)

Now integrating both sides

\(\int (dL)/(L-a)= \int (kx+bk)dx\) \(\ln(L-a)= (kx^2)/(2+bkx+c)\) (because \(\int (1/(x+a)dx=\ln(x+a))\) and \(\int x^n dx= x^{(n+1)/(n+1)}\) where c is the constant of integration

So, the solution of the differential equation \(L'(x) = k(x+b)(L-a)\) will be \(\ln(L-a)= (kx^2)/(2 + bkx + c)\)