Answer
$s=\dfrac{t^3}{3(t-1)^4}-\dfrac{t}{(t-1)^4}+\dfrac{c}{(t-1)^4}$
Work Step by Step
The standard form of the given equation is:
$s'+\dfrac{4}{(t-1)}s=\dfrac{(t+1)}{(t-1)^3}$ ....(1)
The integrating factor is: $e^{\int (4/(t-1)) dx}=(t-1)^4$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [s(t-1)^4]' dt=\int \dfrac{(t+1)}{(t-1)^3}(t-1)^4 dt$
or, $s(t-1)^4=\dfrac{t^3}{3}-t+c$
Thus, the general solution is as follows:
$s=\dfrac{t^3}{3(t-1)^4}-\dfrac{t}{(t-1)^4}+\dfrac{c}{(t-1)^4}$
