Tentukanlah Nilai Integral Tertentu berikut

[tex]\displaystyle \int\limits^3_1\frac{1-\ln x}{x^2}\.dx=\int\limits^3_1(1-\ln x)\frac{1}{x^2}\.dx\\\int\limits^3_1\frac{1-\ln x}{x^2}\.dx=\left-(1-\ln x)\frac{1}{x}\right|^3_1-\int\limits^3_1-\frac1x\cdot\left(-\frac1x\right)\,dx\\\int\limits^3_1\frac{1-\ln x}{x^2}\.dx=\left-(1-\ln x)\frac{1}{x}\right|^3_1-\int\limits^3_1\frac1{x^2}\,dx\\\int\limits^3_1\frac{1-\ln x}{x^2}\.dx=\frac{\ln x-1}{x}+\left\frac1{x}\right|^3_1\\\int\limits^3_1\frac{1-\ln x}{x^2}\.dx=\left\frac{\ln x}{x}\right|^3_1[/tex]
[tex]\displaystyle \int\limits^3_1\frac{1-\ln x}{x^2}\.dx=\frac{\ln 3}{3}-\frac{\ln 1}{1}\\\int\limits^3_1\frac{1-\ln x}{x^2}\.dx=\frac{\ln 3}{3}-0\\\boxed{\boxed{\int\limits^3_1\frac{1-\ln x}{x^2}\.dx=\frac{1}{3}\ln 3}}[/tex]