Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
16,\\
I = \int\limits_1^a {\frac{{{x^3} - 2\ln x}}{{{x^2}}}dx} = \int\limits_1^a {\left( {x - \frac{{2\ln x}}{{{x^2}}}} \right)dx} = \int\limits_1^a {xdx} - 2\int\limits_1^a {\frac{{\ln x}}{{{x^2}}}dx} \\
{I_1} = \int\limits_1^a {\frac{{\ln x}}{{{x^2}}}dx} \\
\left\{ \begin{array}{l}
u = \ln x\\
v' = \frac{1}{{{x^2}}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = \frac{1}{x}\\
v = - \frac{1}{x}
\end{array} \right.\\
\Rightarrow {I_1} = \mathop {\left. {\frac{{ - \ln x}}{x}} \right|}\nolimits_1^a - \int\limits_1^a {\frac{{ - 1}}{{{x^2}}}dx} = \mathop {\left. {\left( {\frac{{ - \ln x}}{x} - \frac{1}{x}} \right)} \right|}\nolimits_1^a \\
\Rightarrow I = \mathop {\left. {\left( {\frac{{{x^2}}}{2} + \frac{{2\ln x}}{x} + \frac{2}{x}} \right)} \right|}\nolimits_1^a \\
= \left( {\frac{{{a^2}}}{2} + \frac{{2\ln a}}{a} + \frac{2}{a}} \right) - \left( {\frac{{{1^2}}}{2} + \frac{{2\ln 1}}{1} + \frac{2}{1}} \right)\\
= \left( {\frac{{{a^2}}}{2} + \frac{2}{a} - \frac{5}{2}} \right) + \frac{{2\ln a}}{a}\\
\Rightarrow a = 2\\
17,\\
I = \int\limits_0^2 {\frac{{{x^2}}}{{x + 1}}dx} = \int\limits_0^2 {\frac{{\left( {{x^2} - 1} \right) + 1}}{{x + 1}}dx} = \int\limits_0^2 {\left( {\frac{{{x^2} - 1}}{{x + 1}} + \frac{1}{{x + 1}}} \right)dx} \\
= \int\limits_0^2 {\left( {x - 1 + \frac{1}{{x + 1}}} \right)dx} = \mathop {\left. {\left( {\frac{{{x^2}}}{2} - x + \ln \left| {x + 1} \right|} \right)} \right|}\nolimits_0^2 \\
= \left( {\frac{{{2^2}}}{2} - 2 + \ln 3} \right) - \left( {\frac{{{0^2}}}{2} - 0 + \ln 1} \right)\\
= \ln 3\\
\Rightarrow a = 0;\,\,\,\,b = 3\\
\Rightarrow ab = 0
\end{array}\)
