Đề bài: Tính tích phân \ I = \int_0^1 {\frac{{\ln \left( {2{x^2} + 4x + 1} \right)}}{{{{\left( {x + 1} \right)}^3}}}} .dx.
Giải:
Đặt: \ \left\{ \begin{array}{l}
u = \ln \left( {2{x^2} + 4x + 1} \right)\\
dv = \frac{{dx}}{{{{\left( {x + 1} \right)}^3}}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \frac{{4\left( {x + 1} \right)}}{{2{x^2} + 4x + 1}}.dx\\
v = - \frac{1}{{2{{\left( {x + 1} \right)}^2}}}
\end{array} \right.
\ \Rightarrow I = - \frac{{\ln \left( {2{x^2} + 4x + 1} \right)}}{{2{{\left( {x + 1} \right)}^2}}}\left| \begin{array}{l}
1\\
0
\end{array} \right. + 2\int_0^1 {\frac{{dx}}{{\left( {x + 1} \right)\left( {2{x^2} + 4x + 1} \right)}}} = 2J - \frac{{\ln 7}}{8}.
\ J = \int_0^1 {\frac{{dx}}{{\left( {x + 1} \right)\left( {2{x^2} + 4x + 1} \right)}}} = \int_0^1 {\left( {\frac{{2x + 2}}{{2{x^2} + 4x + 1}} - \frac{1}{{x + 1}}} \right)dx} .
\ = \frac{1}{2}\int_0^1 {\frac{{4x + 4}}{{2{x^2} + 4x + 1}}dx} - \int_0^1 {\frac{{dx}}{{x + 1}} = \left( {\frac{1}{2}\ln \left| {2{x^2} + 4x + 1} \right| - \ln \left| {x + 1} \right|} \right)} \left| \begin{array}{l}
1\\
0
\end{array} \right. = \frac{{\ln 7}}{2} - \ln 2.
Vậy \ I = 2J - \frac{{\ln 7}}{8} = 2\left( {\frac{{\ln 7}}{2} - \ln 2} \right) - \frac{{\ln 7}}{8} = \frac{7}{8}\ln 7 - 2\ln 2.
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