Cho tích phân $I = \int\limits_1^2 {\dfrac{{x + \ln x}}{{{{\left( {x + 1} \right)}^3}}}{\rm{d}}x}  = a + b.\ln 2 - c.\ln 3$ với $a,b,c \in R$, tỉ số $\dfrac{c}{a}$ bằng

Đặt $\left\{ \begin{array}{l}u = x + \ln x\\{\rm{d}}v = \dfrac{{{\rm{d}}x}}{{{{\left( {x + 1} \right)}^3}}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = \dfrac{{x + 1}}{x}\,{\rm{d}}x\\v =  - \dfrac{1}{{2{{\left( {x + 1} \right)}^2}}}\end{array} \right.$.

Khi đó $I = \left. { - \dfrac{{x + \ln x}}{{2{{\left( {x + 1} \right)}^2}}}} \right|_1^2 + \int\limits_1^2 {\dfrac{{x + 1}}{x}.\dfrac{1}{{2{{\left( {x + 1} \right)}^2}}}{\rm{d}}x} .$

$ =  - \dfrac{{2 + \ln 2}}{{18}} + \dfrac{1}{8} + \dfrac{1}{2}\int\limits_1^2 {\dfrac{{{\rm{d}}x}}{{x\left( {x + 1} \right)}}} $$=  - \dfrac{{2 + \ln 2}}{{18}} + \dfrac{1}{8} + \dfrac{1}{2}\int\limits_1^2 {\left( {\dfrac{1}{x} - \dfrac{1}{{x + 1}}} \right){\rm{d}}x} .$

$\begin{array}{l} =  - \dfrac{{2 + \ln 2}}{{18}} + \dfrac{1}{8} + \dfrac{1}{2}\left. {\left( {\ln \left| x \right| - \ln \left| {x + 1} \right|} \right)} \right|_1^2\\ = \dfrac{1}{{72}} - \dfrac{1}{{18}}\ln 2 + \dfrac{1}{2}\left( {\ln 2 - \ln 3 + \ln 2} \right)\\ = \dfrac{1}{{72}} + \dfrac{{17}}{{18}}\ln 2 - \dfrac{1}{2}\ln 3\\ = a + b.\ln 2 - c.\ln 3.\end{array}$

Vậy $\left\{ \begin{array}{l}a = \dfrac{1}{{72}}\\b = \dfrac{{17}}{{18}}\\c = \dfrac{1}{2}\end{array} \right. \Rightarrow \,\dfrac{c}{a} = \dfrac{1}{2}:\dfrac{1}{{72}} = 36.$