Answer
$\dfrac{1}{9}$
Work Step by Step
Let us consider that $I=\int_{0}^{\infty} x e^{-3x} \ dx$
Plug in $a=-3x \implies da=-3 \ dx$
Integrate the integral by using limits.
$I=\int_{0}^{\infty} \dfrac{e^{a}\ a}{9} \ da\\=\dfrac{1}{9} \int_{0}^{\infty} e^a a \ da \\=\dfrac{1}{9} (a \ e^a-\int_0^{\infty} e^a da \\=\dfrac{1}{9} (a \ e^a-e^a)\\=\dfrac{1}{9} (-3x e^{-3x}-e^{-3x})\\=\lim\limits_{z \to \infty} \int_0^{\infty} x e^{-3x} \ dx \\=\dfrac{1}{9} \lim\limits_{z \to \infty} [-3x e^{-3x} -e^{-3x}]_0^z\\=\dfrac{1}{9}$
