Answer
$P(t)=12000\ln|1+0.25t| +1000$
$P(3)\approx 7715$
Work Step by Step
$P(t)=\displaystyle \int\frac{3000}{1+0.25t}dt=\left[\begin{array}{ll}
u=1+0.25t & \\
du=0.25dt & dt=4du
\end{array}\right]$
$=3000\displaystyle \cdot 4\int\frac{du}{u} =$ ...Log Rule...
$=1200\ln|1+0.25t| +C$
Given that $P(0)=1000$, we find (solve for ) C
$1200\ln|1+0.25(0)| +C=1000$
$1200\ln(1+0)+C=1000$
$C=1000$
So, $P(t)=1200\ln|1+0.25t| +1000$
$P(3)=1200\ln|1+0.25(3)| +1000$
$=12000\ln 1.75+1000\approx$7715