Put $logx=-t \hspace{0.3cm} =\gt\hspace{0.3cm} x=e^{-t}\hspace{0.5cm} =\gt\hspace{0.5cm} dx=-e^{-t}dt$

when

$x=0,t=\infty, x=1,t=0 $

${5cm}\int_{\infty}^{0}e^{-4t}(-t)^4(-e^{-t})dt=\int_{0}^{\infty}e^{-5t}t^4dt $

Put,

$5t=u\hspace{0.3cm} =\gt\hspace{0.3cm} 5dt=du $

${5cm}\int_{0}^{\infty}e^{-u}(\frac{u}{5})^4\frac{du}{5}=\frac{1}{5^5}\int_{0}^{\infty} e^{-u} u^4 du=\frac{1}{5^5}\sqrt{5}=\frac{4!}{5^5}$