Find maximum distance travelled by a body thrown up in air resistance if the air resistance is linearly dependant on its velocity. The mass of object $m$ is given and so is the relation $F_{\text{resistance}} = kv$. Assume a constant $g$.
This is what I have done till now:
$$F_{\text{resistance}} = kv$$ $$ F_{\text{gravity}} = mg$$ $$F = mg - kv$$ $$m \frac{dv}{dt} = mg - kv$$ $$\frac{{\rm d}v}{g-\frac{kv}{m}} = {\rm d}t$$ Integrating both sides $$-\frac{m}{k}\cdot \log\left(g-\frac{kv}{m}\right) = t + c$$ which simplifies to:
$$v(t) = \frac{mg}{k} - Ce^{\frac{-kt}{m}}$$
Given condition $v(0) = u$ so $C = \frac{mg}{k} - u$ and
$$v(t) = \frac{mg}{k} - \left(\frac{mg}{k} - u\right)e^{\frac{-kt}{m}}$$
Let max height be $x$ and time taken to reach this height be $T$. Since at the top, $v(T) = 0$,
therefore,
$$\frac{mg}{k} = \left(\frac{mg}{k} -u\right)e^{-\frac{kT}{m}}$$ or $$\log(mg/k)= \log(mg/k -u) - kT/m$$ $$kT/m = \log((mg -ku)/mg)$$ $$T = (m/k)\log(1 - ku/mg)$$
I need to prove that if $ku/mg <1$
$$T = u/g[1- (1/2)(ku/mg) + 1/3(ku/mg)^2 + \ldots]$$ and $$X = u^2/2g[1 - (2/3)(ku/mg) + 1/2(ku/mg)^2 + \ldots]$$
I don't know what to do now. I tried integrating $v(t)$ from $0$ to $T$ to get $X$ but thats such a complicated form and nothing similar to the required equation. All help is greatly Appreciated!
Take upward as positve
\begin{align*} m\frac{dv}{dt} &= -mg-kv \\ \frac{dv}{v+\frac{mg}{k}} &= -\frac{k}{m} \, dt \\ \ln \left( v+\frac{mg}{k} \right)&=-\frac{kt}{m}+C \\ v &= \left( u+\frac{mg}{k} \right)e^{-\frac{kt}{m}}-\frac{mg}{k} \\ s &= \frac{m}{k} \left( u+\frac{mg}{k} \right) \left( 1-e^{-\frac{kt}{m}} \right)-\frac{mgt}{k} \end{align*}
At maximal height,
\begin{align*} v &= 0 \\ t &= \frac{m}{k} \ln \left( 1+\frac{ku}{mg} \right) \\ s &= \frac{mu}{k}-\frac{m^2 g}{k^2} \ln \left( 1+\frac{ku}{mg} \right) \\ & \approx \frac{u^2}{2g}-\frac{ku^3}{3mg^2}+\frac{k^2u^4}{4m^2g^3} \end{align*}
where $0<\dfrac{ku}{mg}<1$