Ta có a + b = 5 .

d) Đúng.Ta có \(\int\limits_1^3 {\frac{{xf\left( x \right) + {x^2} - 1}}{x}} \,{\rm{d}}x = \int\limits_1^3 {\left[ {f\left( x \right) + x - \frac{1}{x}} \right]} \,{\rm{d}}x = \int\limits_1^3 {f\left( x \right)} \,{\rm{d}}x + \int\limits_1^3 x {\rm{d}}x - \int\limits_1^3 {\frac{1}{x}} \,{\rm{d}}x\)

                    \( = 2 + \left. {\frac{{{x^2}}}{2}} \right|_1^3 - \left. {\ln \left| x \right|} \right|_1^3 = 2 + \frac{9}{2} - \frac{1}{2} - \ln 3 + \ln 1 = 6 - \ln 3 = a + b\ln 3\).

Suy ra \(a = 6;b =  - 1 \Rightarrow a + b = 5\).