You differentiate both sides of the equation given above with respect to t. Differentiating 1/t yields -1/t^2. On the LHS you apply the differentiation inside the integral.

∫0->∞ x*e^-tx dx = -∂_t ∫0->∞ e^-tx dx = -∂_t (1/t)∫0->∞ e^-u du = -∂_t (1/t) = 1/t^2