Đáp án:
\[T = \frac{{152}}{{27}}\]
Giải thích các bước giải:
\(\begin{array}{l}
I = \int\limits_1^2 {\frac{{\ln x}}{{{{\left( {x + 1} \right)}^2}}}dx} \\
\left\{ \begin{array}{l}
u = \ln x\\
v' = \frac{1}{{{{\left( {x + 1} \right)}^2}}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = \frac{1}{x}\\
v = \frac{{ - 1}}{{x + 1}}
\end{array} \right.\\
\Rightarrow I = \mathop {\left. {\frac{{\left( { - 1} \right).\ln x}}{{x + 1}}} \right|}\nolimits_1^2 - \int\limits_1^2 {\frac{1}{x}.\frac{{ - 1}}{{x + 1}}dx} \\
= \frac{{ - \ln 2}}{{2 + 1}} - \frac{{ - \ln 1}}{{1 + 2}} + \int\limits_1^2 {\frac{{\left( {x + 1} \right) - x}}{{x\left( {x + 1} \right)}}dx} \\
= - \frac{{\ln 2}}{3} + \int\limits_1^2 {\left[ {\frac{1}{x} - \frac{1}{{x + 1}}} \right]dx} \\
= - \frac{{\ln 2}}{3} + \mathop {\left. {\left( {\ln x - \ln \left( {x + 1} \right)} \right)} \right|}\nolimits_1^2 \\
= - \frac{{\ln 2}}{3} + \left( {\ln 2 - \ln 1 - \ln 3 + \ln 2} \right)\\
= - \frac{1}{3}\ln 2 + 2\ln 2 - \ln 3\\
= - \ln 3 + \frac{5}{3}\ln 2\\
\Rightarrow \left\{ \begin{array}{l}
a = - 1\\
b = \frac{5}{3}
\end{array} \right. \Rightarrow T = {a^2} + {b^3} = \frac{{152}}{{27}}
\end{array}\)
