## Calculus (3rd Edition)

We have $\int_0^1\dfrac{\ln x dx}{x^2} \ dx=\int_0^{1}\dfrac{1}{x^2} \ln x \ dx\\ =-\dfrac{1}{x}\ln x+\int\dfrac{dx}{x^2}\\=-\dfrac{\ln x }{x}-\dfrac{1}{x} +C$ We evaluate the integral with the limit as $R\to0^{+}$ $\int_0^1\dfrac{\ln x dx}{x^2} \ dx=\lim\limits_{R \to 0^{+}}(\dfrac{1}{R}\ln R+\dfrac{1}{R}-1)\\=-1+\lim\limits_{R \to 0^{+}}(\dfrac{\ln R+1}{R})=\infty$ Hence, the integral diverges.