If $\lambda=\lambda\left(r\right)$ , how do you get from \left(1-2r\lambda'\right)e^{-2\lambda}=1 (the textbook then says, “integrates to”)$\frac{d\left(re^{-2\lambda}\right)}{dr}=1$
$re^{-2\lambda}=r+c$
I can see how you get from the second to the third equation, but not how to get from the first to the second? The prime means differentiated wrt r.
Many thanks.