Tính tích phân 1 to e (lnx/x^2)dx

 $I = \int\limits_1^e {\frac{{\ln x}}{{{x^2}}}dx}$ Đặt $\left\{ \begin{array}{l} \ln x = u \to du = \frac{1}{x}dx\\ \frac{{dx}}{{{x^2}}} = dv \Rightarrow v = - \frac{1}{x} \end{array} \right.$

Khi đó $I = \left. { - \frac{1}{x}.\ln x} \right|_1^e - \int\limits_1^e { - \frac{1}{x}.\frac{1}{x}dx}$

$= \left( { - \frac{1}{e}.{\mathop{\rm lne}\nolimits} } \right) - \left( { - \frac{1}{1}.\ln 1} \right) + \int\limits_1^e {\frac{1}{{{x^2}}}dx}$

$= \frac{{ - 1}}{e} + \left. {\left( { - \frac{1}{x}} \right)} \right|_1^e = \frac{{ - 1}}{e} - \frac{1}{e} + \frac{1}{1} = 1 - \frac{2}{e}$

Vậy a=1; b=-2 nên a+b=-1.