Show that the distance traveled by the particle at any time t is $x=\left(\frac{g}{k^2}+\frac{u}{k}\right)(1-e^{-kt})-\frac{g}{k}t$

$ \frac{dv}{dt}=-g-kv\\ \frac{dv}{g+kv}=-dt$

Integrating the above equation we get,

$\frac{1}{k}\log (g+kv)=-t+c$

$Initially\ when\ t=0,\ v=u$

$\frac{1}{k}(g+ku)=c$

$\frac{1}{k}\log (g+kv)=-t+\frac{1}{k}\log(g+ku)$

$ t=\frac{1}{k}\log\frac{(g+ku)}{(g+kv)}$

$kt=\log\frac{(g+ku)}{(g+kv)} $

$ e^{kt}=\frac{(g+ku)}{(g+kv)}$

$e^{-kt}(g+ku)=(e+gv)$

$-g+(g+ku)e^{-kt}=(kv)$

$v=-\frac{g}{k}+\frac{1}{k}(g+ku)e^{-kt}$

$ \frac{dx}{dt}=\frac{-g}{k}+\frac{1}{k}(g+ku)e^{-kt} $

Integrating the above equation we get,

$x=-\frac{g}{k}t-\frac{1}{k^2}(g+ku)e^{-kt}+c $

Initially when $t=0,x=0$ we get $c=\frac{(g+ku)}{k^2}$

$ x=-\frac{g}{k}tt-\frac{1}{k^2}(g+ku)e^{-kt}+\frac{(g+ku)}{k^2}$

$x= \frac{(g+ku)}{k^2}(1-e^{-kt})-\frac{g}{k}t$

$x=\left( \frac{g}{k^2}+\frac{u}{k}\right)(1-e^{-kt})-\frac{g}{k}t$