Tính \(\displaystyle \int\limits_0^1 {\frac{{x + 2}}{{{x^2} + 2x + 1}}\ln (x + 1)dx} \)
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Lời giải:
Báo saiTa có: \(\displaystyle \frac{{x + 2}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{x + 1}}{{{{\left( {x + 1} \right)}^2}}} + \frac{1}{{{{\left( {x + 1} \right)}^2}}}\) \(\displaystyle = \frac{1}{{x + 1}} + \frac{1}{{{{\left( {x + 1} \right)}^2}}}\)
Khi đó \(\displaystyle \int\limits_0^1 {\frac{{x + 2}}{{{x^2} + 2x + 1}}\ln (x + 1)dx} \)\(\displaystyle = \int\limits_0^1 {\frac{{\ln \left( {x + 1} \right)}}{{x + 1}}dx} + \int\limits_0^1 {\frac{{\ln \left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}dx} \) \(\displaystyle = I + J\)
\(\displaystyle I = \int\limits_0^1 {\ln \left( {x + 1} \right)d\left( {\ln \left( {x + 1} \right)} \right)} \) \(\displaystyle = \left. {\frac{{{{\ln }^2}\left( {x + 1} \right)}}{2}} \right|_0^1 = \frac{{{{\ln }^2}2}}{2}\)
Tính \(\displaystyle J = \int\limits_0^1 {\frac{{\ln \left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}dx} \).
Đặt \(\displaystyle \left\{ \begin{array}{l}u = \ln \left( {x + 1} \right)\\dv = \frac{{dx}}{{{{\left( {x + 1} \right)}^2}}}\end{array} \right.\) \(\displaystyle \Rightarrow \left\{ \begin{array}{l}du = \frac{1}{{x + 1}}dx\\v = - \frac{1}{{x + 1}}\end{array} \right.\)
\(\displaystyle \Rightarrow J = - \left. {\frac{{\ln \left( {x + 1} \right)}}{{x + 1}}} \right|_0^1 + \int\limits_0^1 {\frac{1}{{{{\left( {x + 1} \right)}^2}}}dx} \) \(\displaystyle = - \frac{{\ln 2}}{2} - \left. {\frac{1}{{x + 1}}} \right|_0^1\) \(\displaystyle = - \frac{{\ln 2}}{2} - \frac{1}{2} + 1 = \frac{1}{2} - \frac{{\ln 2}}{2}\)
Vậy \(\displaystyle \int\limits_0^1 {\frac{{x + 2}}{{{x^2} + 2x + 1}}\ln (x + 1)dx} \)\(\displaystyle = \frac{{{{\ln }^2}2}}{2} + \frac{1}{2} - \frac{{\ln 2}}{2}\) \(\displaystyle = \frac{{{{\ln }^2}2 - \ln 2 + 1}}{2}\)
