Answer
$\frac{1}{5}t^5 \ln t- \frac{1}{25}t^5 +C$
Work Step by Step
Let $u = ln t, dv = t^4 dt$ and $du = \frac{1}{t} dt, v = \frac{1}{5}t^5$
Then
$\int t^4 \ln t dt$
$=\frac{1}{5} t^5 \ln t -\int \frac{1}{5} t^5 \frac{1}{t} dt$
$=\frac{1}{5} t^5 \ln t -\int \frac{1}{5} t^4 dt$
$=\frac{1}{5}t^5 \ln t- \frac{1}{25}t^5 +C$
