Correction exercice I
Calcule de l’intégrale :
\(I = \) \(\int {\frac{{(x - 1)dx}}{{{{\left( {{x^2} + 2x + 3} \right)}^2}}}} \) \( = \) \(\int {\frac{{\frac{1}{2}(2x + 2) + ( - 1 - 1)}}{{{{\left( {{x^2} + 2x + 3} \right)}^2}}}} \) \(dx = \frac{1}{2}\) \(\int {\frac{{(2x + 2)}}{{{{\left( {{x^2} + 2x + 3} \right)}^2}}}} \) \(dx - 2\) \(\int {\frac{{dx}}{{{{\left( {{x^2} + 2x + 3} \right)}^2}}}} \)
\(I = - \frac{1}{2}\) \(\frac{1}{{\left( {{x^2} + 2x + 3} \right)}}\) \( - 2\) \(\int {\frac{{dx}}{{{{\left( {{x^2} + 2x + 3} \right)}^2}}}} \)
\(\int {\frac{{dx}}{{{{\left( {{x^2} + 2x + 3} \right)}^2}}}} \) \( = \) \(\int {\frac{{dx}}{{{{\left[ {{{\left( {x + 1} \right)}^2} + 2} \right]}^2}}}} \) \( = \int {\frac{{dt}}{{{{\left( {{t^2} + 2} \right)}^2}}}} \) \( = \frac{1}{2}\) \(\int {\frac{{({t^2} + 2) - {t^2}}}{{{{\left( {{t^2} + 2} \right)}^2}}}} \) \(dt = \frac{1}{2}\) \(\int {\frac{{dt}}{{{t^2} + 2}}} \) \( - \frac{1}{2}\) \(\int {\frac{{{t^2}dt}}{{{{\left( {{t^2} + 2} \right)}^2}}}} \) ayant posé \(t = x + 2\)
\(\int {\frac{{{t^2}dt}}{{{{\left( {{t^2} + 2} \right)}^2}}}} \) \( = - \frac{1}{2}\) \(\frac{t}{{\left( {{t^2} + 2} \right)}}\) \( + \frac{1}{{2\sqrt 2 }}\) \(\arctan \frac{t}{{\sqrt 2 }}\)
\(I = - \) \(\frac{{x + 2}}{{2\left( {{x^2} + 2x + 3} \right)}}\) \( - \frac{{\sqrt 2 }}{4}\) \(\arctan \left( {\frac{{x + 2}}{{\sqrt 2 }}} \right)\) \( + cte\)
b) \(I = \) \(\int {\frac{{xdx}}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}} \)
En décomposant I en éléments simples, nous avons :
\(\frac{x}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}\) \( = \frac{{Ax + B}}{{{x^2} + 1}}\) \( + \frac{C}{{x - 1}}\)
Nous retrouvons donc, après développement et réduction au même dénominateur \(\left\{ \begin{array}{l}A = - \frac{1}{2}\\B = \frac{1}{2}\\C = \frac{1}{2}\end{array} \right.\)
\(I = \) \( - \frac{1}{2}\int {\frac{{x - 1}}{{{x^2} + 1}}} dx\) \( + \frac{1}{2}\int {\frac{{dx}}{{x - 1}}} \)
\(I = - \frac{1}{4}\) \(Log\left| {{x^2} + 1} \right| + \) \(\frac{1}{2}\arctan x + \) \(\frac{1}{2}Log\left| {x - 1} \right|\) \( + cte\)
c) \(I = \) \(\int {\frac{{{x^4} + 4{x^3} + 11{x^2} + 12x + 8}}{{{{\left( {{x^2} + 2x + 3} \right)}^2}\left( {x + 1} \right)}}} \) \(dx\)
En décomposant I en éléments simples, nous avons :
\(\frac{{{x^4} + 4{x^3} + 11{x^2} + 12x + 8}}{{{{\left( {{x^2} + 2x + 3} \right)}^2}\left( {x + 1} \right)}}\) \( = \) \(\frac{{Ax + B}}{{{{\left( {{x^2} + 2x + 3} \right)}^2}}}\) \( + \frac{{Cx + D}}{{\left( {{x^2} + 2x + 3} \right)}}\) \( + \frac{E}{{x + 1}}\)
Vous retrouvez \(\left\{ \begin{array}{l}A = 1\\B = - 1\\C = 0\end{array} \right.\) et \(\left\{ \begin{array}{l}D = 0\\E = 1\end{array} \right.\)
\(I = \) \(\int {\frac{{x - 1}}{{{{\left( {{x^2} + 2x + 3} \right)}^2}}}dx} \) \( + \int {\frac{{dx}}{{x + 1}}} \)
\(I = - \) \(\frac{{x + 2}}{{2\left( {{x^2} + 2x + 3} \right)}}\) \( - \frac{{\sqrt 2 }}{4}\arctan \frac{{x + 1}}{{\sqrt 2 }}\) \( + Log\left| {x + 1} \right|\) \( + cte\)
