Answer
$\displaystyle L= -\frac{1}{kt\ln(t)-kt+1+k}$
Work Step by Step
$\displaystyle \frac{dL}{dt} = kL^2\ln t$
Separation of Variables
$\displaystyle \frac{dL}{L^2} = k\ln (t)dt$
Integrate with respect to each variable. To integrate $\ln(t)$ we have to use integration by parts whose formula is the following (Let $f$ and $g$ be functions of $x$):$\displaystyle \int (f\cdot g')dx = f\cdot g-\int (f'\cdot g)dx$
For this problem, let $f = \ln(t)$ such that $f' = \frac{1}{t}$ and $g' = 1$ such that $g = t$
$\displaystyle \int\frac{dL}{L^2} = k\int(1)\ln (t)dt$
$\displaystyle \int L^{-2}dL= k(t\ln(t)-\int t\cdot\frac{1}{t}dt)$
$\displaystyle \frac{L^{-1}}{-1}= k(t\ln(t)-\int (1)dt)$
$\displaystyle -\frac{1}{L}= k(t\ln(t)-t+C)$
Plug in the initial condition to solve for $C$
$\displaystyle -\frac{1}{(-1)} = k((1)\ln(1)-(1)+C)$
$\displaystyle 1 = k(0-1+C)$
$\displaystyle \frac{1}{k} = -1+C$
$\displaystyle \frac{1}{k} + 1 = C$
Plug the value for $C$ that we just found back into the equation and isolate $L$
$\displaystyle -\frac{1}{L}= k(t\ln(t)-t+\frac{1}{k}+1) = kt\ln(t)-kt+1+k$
$\displaystyle \frac{1}{L}= -(kt\ln(t)-kt+1+k)$
$\displaystyle L= -\frac{1}{kt\ln(t)-kt+1+k}$
