Answer
$\infty$, diverges
Work Step by Step
$\int_{1}^{\infty}x^{2}lnxdx$
$\lim\limits_{a \to \infty}\int_{1}^{a}x^{2}lnxdx$
Perform integration by parts, where the integral is equal to $uv-\int(vdu)$:
$u=lnx$
$du=\frac{1}{x}dx$
$\int(v)=\int(x^{2})$
$v=\frac{1}{3}x^{3}$
$\lim\limits_{a \to \infty}[(\frac{1}{3}x^{3})(lnx)-\int(\frac{1}{3}x^{3})(\frac{1}{x})dx]_{1}^{a}$
$\lim\limits_{a \to \infty}[\frac{1}{3}x^{3}lnx-\int(\frac{1}{3}x^{2}))dx]_{1}^{a}$
$\lim\limits_{a \to \infty}[\frac{1}{3}x^{3}lnx-\frac{1}{9}x^{3}]_{1}^{a}$
$\lim\limits_{a \to \infty}[\frac{1}{3}a^{3}lna-\frac{1}{9}a^{3}-\frac{1}{3}(1)^{3}ln(1)+\frac{1}{9}(1)^{3}]$
$\lim\limits_{a \to \infty}[\frac{1}{3}a^{3}lna-\frac{1}{9}a^{3}+\frac{1}{9}]$=$\infty-\infty$
$\infty-\infty$ is indeterminate, so we must manipulate the function:
$\lim\limits_{a \to \infty}[\frac{1}{3}a^{3}(lna-\frac{1}{3})+\frac{1}{9}]$=$\infty\times\infty$=$\infty$
