Answer
$= (\frac{1}{5})t^{5}(\ln t) - \frac{1}{25}t^{5} +C$
Work Step by Step
$\int t^{4} \ln(t) dt$
$u = \ln t$
$u' = \frac{1}{t}$
$\frac{du}{dt} = \frac{1}{t}$
$du = \frac{dt}{t}$
$dv = t^{4}$
$v = \frac{t^{5}}{5}$
$uv - \int vdu$
$= (\ln t)(\frac{t^{5}}{5}) - \int \frac{t^{5}}{5} \frac{dt}{t}$
$= (\ln t)(\frac{t^{5}}{5}) - \int \frac{t^{5}}{5} \frac{1}{t}dt$
$= (\ln t)(\frac{t^{5}}{5}) - \int \frac{t^{5}}{5t}dt$
$= (\ln t)(\frac{t^{5}}{5}) - \int \frac{t^{4}}{5}dt$
$= (\ln t)(\frac{t^{5}}{5}) - \frac{1}{5}\int t^{4}dt$
$= (\ln t)(\frac{t^{5}}{5}) - \frac{1}{5}\frac{t^{5}}{5}$
$= (\ln t)(\frac{t^{5}}{5}) - \frac{t^{5}}{25} +C$
$= (\frac{1}{5})t^{5}(\ln t) - \frac{1}{25}t^{5} +C$
