Answer: \(x=\dfrac{2t+1}{2t-1}\left(t+\ln\left(\dfrac{3}{2t+1}\right)\right)\).
First put the equation into linear form by dividing through by \(4t^2-1\):
\(\dfrac{dx}{dt}+\dfrac{4}{4t^2-1}x=1\).
Use partial fractions:
\(\dfrac{4}{4t^2-1}=\dfrac{4}{(2t-1)(2t+1)}=\dfrac{2}{2t-1}-\dfrac{2}{2t+1}\).
An integrating factor is therefore
\(\exp\left(\int \dfrac{4}{4t^2-1}\,dt\right)=\exp\left(\ln(2t-1)-\ln(2t+1)\right)=\dfrac{2t-1}{2t+1}\).
Multiplying the differential equation by this factor gives
\(\dfrac{d}{dt}\left(x\dfrac{2t-1}{2t+1}\right)=\dfrac{2t-1}{2t+1}\).
Now integrate the right hand side:
\(\dfrac{2t-1}{2t+1}=1-\dfrac{2}{2t+1}\),
so
\(\int \dfrac{2t-1}{2t+1}\,dt=t-\ln(2t+1)\).
Hence
\(x\dfrac{2t-1}{2t+1}=t-\ln(2t+1)+C\).
Use \(x=3\) when \(t=1\):
\(3\cdot\dfrac{1}{3}=1-\ln 3+C\).
Thus \(1=1-\ln 3+C\), so \(C=\ln 3\).
Therefore
\(x\dfrac{2t-1}{2t+1}=t-\ln(2t+1)+\ln 3=t+\ln\left(\dfrac{3}{2t+1}\right)\).
Finally,
\(x=\dfrac{2t+1}{2t-1}\left(t+\ln\left(\dfrac{3}{2t+1}\right)\right)\).
