Let u=x/t, then du/dt=-x/t²+(1/t)dx/dt, so dx/dt=tdu/dt+x/t=tdu/dt+u.
So we can write dx/dt=tdu/dt+u=u+1/u+1, tdu/dt=1/u+1=(1+u)/u.
Therefore ∫(u/(1+u)du=∫dt/t, ∫(1-1/(1+u))du=ln(at) where a is constant of integration.
ln(at)=u-ln(1+u)=(x/t)-ln(1+(x/t)).