$\dfrac{ds}{dt}=\dfrac{2s^2}{s^2+t^2}$
$\dfrac{dt}{ds}=\dfrac{s^2+t^2}{2s^2}$
$\dfrac{dt}{ds}=\dfrac{1}{2}\left(1+\left(\dfrac{t}{s}\right)^2\right)$
Let $u=\dfrac{t}{s}$ ,
Then $t=su$
$\dfrac{dt}{ds}=s\dfrac{du}{ds}+u$
$\therefore s\dfrac{du}{ds}+u=\dfrac{1+u^2}{2}$
$s\dfrac{du}{ds}=\dfrac{u^2+1}{2}-u$
$s\dfrac{du}{ds}=\dfrac{u^2-2u+1}{2}$
$s\dfrac{du}{ds}=\dfrac{(u-1)^2}{2}$
$\dfrac{du}{(u-1)^2}=\dfrac{ds}{2s}$
$\int\dfrac{du}{(u-1)^2}=\int\dfrac{ds}{2s}$
$-\dfrac{1}{u-1}=\dfrac{\ln s}{2}+c$
$\dfrac{1}{u-1}=\dfrac{C-\ln s}{2}$
$u=\dfrac{2}{C-\ln s}+1$
$\dfrac{t}{s}=\dfrac{2}{C-\ln s}+1$
$t=\dfrac{2s}{C-\ln s}+s$
$s(0)=1$ :
$0=\dfrac{2}{C}+1$
$C=-2$
$\therefore t=s-\dfrac{2s}{\ln s+2}$