Simply write your equation as \begin{align} \frac{dR}{kR(1300-R)} = dt \end{align} and integrate! [I figure out you did typo on your notes!]
And then managing $t$ then you will get the answer.
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About @deviljones comment, try $x=\frac{1}{R}$[since $R=R(t)$, $x=x(t)$], then \begin{align} -\frac{dx}{1300kx-k} = dt \end{align} after integration you have \begin{align} ln(k-1300kx) = -1300kt + C \end{align} Then expressing in terms of $x=\frac{1}{R}$, you get what you want.