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[ID: 1806] [Date de publication: 12 mai 2021 12:10] [Catégorie(s): Fractions rationnelles ] [ Nombre commentaires: 1] [nombre d'éditeurs: 1 ] [Editeur(s): Emmanuel Vieillard-Baron ] [nombre d'auteurs: 3 ] [Auteur(s): Emmanuel Vieillard-Baron Alain Soyeur François Capaces ]

Solution(s)

Solution(s)

Exercice 88
Par Emmanuel Vieillard-Baron Alain Soyeur François Capaces le 12 mai 2021 12:10

1. \({\scriptstyle 1\over\scriptstyle x\left(x-1\right)} = {\scriptstyle-1\over\scriptstyle x} + {\scriptstyle 1\over\scriptstyle x-1}\) donc \(\int_{2}^{3} {\scriptstyle 1\over\scriptstyle x\left(x-1\right)}\,\textrm{d}x = \Bigl[ -\ln\left|x\right| + \ln\left|x-1\right| \Bigr]_{2}^{3}=\ln \left({\scriptstyle 4\over\scriptstyle 3}\right)\)

2. \(\int_{2}^{3} {\scriptstyle 2x+1\over\scriptstyle x^2-1}\,\textrm{d}x = {\scriptstyle 3\over\scriptstyle 2} \int_{2}^{3} {\scriptstyle 1\over\scriptstyle x-1}\,\textrm{d}x + \int_{1}^{2} {\scriptstyle 1\over\scriptstyle x+1}\,\textrm{d}x={\scriptstyle 1\over\scriptstyle 2}\Bigl[ 3\ln \left|x-1\right|+ \ln\left|x+1\right| \Bigr]_{2}^{3}={\scriptstyle 1\over\scriptstyle 2}\left(3\ln 2 + \ln 4 - \ln 3\right)\)

3. \(\int_{0}^{1} {\scriptstyle x^2-1\over\scriptstyle x^2+4x-5}\,\textrm{d}x = \int_{0}^{1} \left({\scriptstyle x+1\over\scriptstyle x+5}\right)\,\textrm{d}x = \int_{0}^{1} 1-{\scriptstyle 4\over\scriptstyle x+5}\,\textrm{d}x=\Bigl[ x-4\ln \left|x+5\right| \Bigr]_{0}^{1} = 1-4\ln 6 + 4\ln 5\).

4. \(\int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle x+1\over\scriptstyle\left(x^2+1\right)\left(x-2\right)}\,\textrm{d}x={\scriptstyle 3\over\scriptstyle 5} \int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle 1\over\scriptstyle x-2}\,\textrm{d}x - {\scriptstyle 3\over\scriptstyle 5} \int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle x+{\scriptstyle 1\over\scriptstyle 2}\over\scriptstyle x^2+1}\,\textrm{d}x = \Bigl[ {\scriptstyle 3\over\scriptstyle 5}\ln \left|x-2\right| \Bigr]_{0}^{{\scriptstyle 1\over\scriptstyle 2}} - {\scriptstyle 3\over\scriptstyle 5} \int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle x+{\scriptstyle 1\over\scriptstyle 2}\over\scriptstyle x^2+1}\,\textrm{d}x= {\scriptstyle 3\over\scriptstyle 5}\ln \left({\scriptstyle 3\over\scriptstyle 4}\right) -{\scriptstyle 3\over\scriptstyle 5}\int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle x\over\scriptstyle x^2+1}\,\textrm{d}x - {\scriptstyle 1\over\scriptstyle 5}\int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle 1\over\scriptstyle x^2+1}\,\textrm{d}x ={\scriptstyle 3\over\scriptstyle 5}\ln \left({\scriptstyle 3\over\scriptstyle 4}\right) + \Bigl[ -{\scriptstyle 3\over\scriptstyle 10} \ln \left|x^2+1\right| - {\scriptstyle 1\over\scriptstyle 5}\operatorname{arctan} x \Bigr]_{0}^{{\scriptstyle 1\over\scriptstyle 2}}={\scriptstyle 3\over\scriptstyle 5} \ln ({\scriptstyle 3\over\scriptstyle 4}) - {\scriptstyle 3\over\scriptstyle 10} \ln({\scriptstyle 5\over\scriptstyle 4}) - {\scriptstyle 1\over\scriptstyle 5} \operatorname{arctan} ({\scriptstyle 1\over\scriptstyle 2})\)

5. On écrit la décomposition a priori \(\dfrac{1}{(x+2)(x^2+2x+5)} = \dfrac{A}{x-2}+\dfrac{Bx+C}{x^2+2x+5}\).
   En multipliant les deux membres par \(x+2\) et en faisant \(x=-2\) on trouve \(A = \dfrac{1}{4-4+5} = \dfrac15\).
   En multipliant les deux membres par \(x^2+2x+5\) et en faisant \(x=-1+2i\) on trouve \(B(-1+2i)+C = \dfrac{1}{-1+2i+2} = \dfrac{1-2i}{5}\). D’où \(B = -\dfrac15\) et \(C = 0\).
   \[\begin{aligned} \int_0^1 \dfrac{\textrm dx}{(x+2)(x^2+2x+5)} &= \dfrac15 \int_0^1 \dfrac{\textrm dx}{x+2} - \dfrac15\int_0^1 \dfrac{\textrm dx}{x^2+2x+5} \\ &= \dfrac1{10} \left[ 2\ln(x+2) - \ln(x^2+2x+5) \right]_0^1 \\ &= \dfrac1{10} \left( 2\ln3 - 2\ln2 -\ln8 + \ln5 \right) \\ &= \dfrac1{10} \ln \left( \dfrac{45}{32}\right).\end{aligned}\]

6. On écrit la décomposition a priori \(\dfrac{x}{(x+1)(x^2+x+1)} = \dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+x+1}\).
   En multipliant les deux membres par \(x+1\) et en faisant \(x=-1\) on trouve \(A = \dfrac{-1}{1-1+1} = -1\).
   En multipliant les deux membres par \(x^2+x+1\) et en faisant \(x=j\) on trouve \(Bj+C = \dfrac{j}{j+1} = \dfrac{j(j^2+1)}{(j+1)(j^2+1)} = 1+j\). D’où \(B = 1\) et \(C = 1\).
   \[\begin{aligned} \int_0^2 \dfrac{x\,\textrm dx}{(x+1)(x^2+x+1)} &= -\int_0^2 \dfrac{\textrm dx}{x+1} + \dfrac12 \int_0^2 \dfrac{2x+1}{x^2+x+1}\,\textrm dx \dfrac12 \int_0^2 \dfrac{\textrm dx}{x^2+x+1} \\ &= \left[ -\ln (x+1)+ \dfrac12 \ln(x^2+x+1) \right]_0^2 + \dfrac12\int_0^2 \dfrac{\textrm dx}{(x+{\scriptstyle 1\over\scriptstyle 2})^2+{\scriptstyle 3\over\scriptstyle 4}} \\ &= -\ln3+ \dfrac12 \ln7 + \dfrac12\times\dfrac43 \int_0^2 \dfrac{\textrm dx}{({\scriptstyle 2x+1\over\scriptstyle\sqrt 3})^2+1}\end{aligned}\] On pose \(u = {\scriptstyle 2x+1\over\scriptstyle\sqrt 3}\), \(\textrm du = \dfrac{2\,\textrm dx}{\sqrt3}\), \[\begin{aligned} \dfrac12\times\dfrac43 \int_0^2 \dfrac{\textrm dx}{({\scriptstyle 2x+1\over\scriptstyle\sqrt 3})^2+1} &= \dfrac{\sqrt3}{6}\int_{{\scriptstyle 1\over\scriptstyle\sqrt 3}}^{{\scriptstyle 5\over\scriptstyle\sqrt 3}} \dfrac{\textrm du}{u^2+1}\\ &= \dfrac{\sqrt3}{6} \left( \operatorname{arctan} \left( {\scriptstyle 5\over\scriptstyle\sqrt 3}\right) - \operatorname{arctan} \left( {\scriptstyle 1\over\scriptstyle\sqrt 3}\right) \right) \end{aligned}\] Maintenant \[\begin{aligned} \tan \left( \operatorname{arctan} \left( {\scriptstyle 5\over\scriptstyle\sqrt 3}\right) - \operatorname{arctan} \left( {\scriptstyle 1\over\scriptstyle\sqrt 3}\right) \right) &= \dfrac{{\scriptstyle 5\over\scriptstyle\sqrt 3} - {\scriptstyle 1\over\scriptstyle\sqrt 3}}{1+{\scriptstyle 5\over\scriptstyle\sqrt 3}\times {\scriptstyle 1\over\scriptstyle\sqrt 3}} \\ &= \dfrac{{\scriptstyle 4\over\scriptstyle\sqrt 3}}{1+{\scriptstyle 5\over\scriptstyle 3}}\\ &= \dfrac{\sqrt3}{2}\end{aligned}\] Donc \(\operatorname{arctan} \left( {\scriptstyle 5\over\scriptstyle\sqrt 3}\right) - \operatorname{arctan} \left( {\scriptstyle 1\over\scriptstyle\sqrt 3}\right) = \operatorname{arctan} \left( \dfrac{\sqrt3}{2}\right) + k\pi\). D’après la propriété des accroissements finis, \(\operatorname{arctan} \left( {\scriptstyle 5\over\scriptstyle\sqrt 3}\right) - \operatorname{arctan} \left( {\scriptstyle 1\over\scriptstyle\sqrt 3}\right) \leqslant {\scriptstyle 5\over\scriptstyle\sqrt 3} - {\scriptstyle 1\over\scriptstyle\sqrt 3}\),
   donc \(\left\vert \operatorname{arctan} \left( {\scriptstyle 5\over\scriptstyle\sqrt 3}\right) - \operatorname{arctan} \left( {\scriptstyle 1\over\scriptstyle\sqrt 3}\right) - \operatorname{arctan} \left( \dfrac{\sqrt3}{2}\right) \right\vert \leqslant \dfrac{4}{\sqrt3} + \dfrac{\sqrt3}{2} < \pi\) d’où \(k= 0\) et \[\int_0^2 \dfrac{x\,\textrm dx}{(x+1)(x^2+x+1)} = -\ln3+ \dfrac12 \ln7 + \dfrac{\sqrt3}{6} \operatorname{arctan} \left( \dfrac{\sqrt3}{2}\right).\]

7. \[\begin{aligned} \int_0^1 \dfrac{\textrm dx}{(x^2+x+1)^2} &= \int_0^1 \dfrac{\textrm dx}{\left( (x+{\scriptstyle 1\over\scriptstyle 2})^2+{\scriptstyle 3\over\scriptstyle 4}\right)^2 } \\ &= \dfrac{16}{9} \int_0^1 \dfrac{\textrm dx}{\left( ({\scriptstyle 2x+1\over\scriptstyle\sqrt 3})^2+1\right)^2 } \\ &= \dfrac{16}{9} \times \dfrac{\sqrt3}{2}\int_{{\scriptstyle 1\over\scriptstyle\sqrt 3}}^{{\scriptstyle 3\over\scriptstyle\sqrt 3}} \dfrac{\textrm du}{(u^2+1)^2}\end{aligned}\] En posant \(u = \tan\varphi,\,\textrm d\varphi= \dfrac{\textrm du}{u^2+1}\), \[\begin{aligned} \int_0^1 \dfrac{\textrm dx}{(x^2+x+1)^2} &= \dfrac{8\sqrt3}{9} \int_{\operatorname{arctan} \left( {\scriptstyle 1\over\scriptstyle\sqrt 3}\right) }^{\operatorname{arctan} \left({\scriptstyle 3\over\scriptstyle\sqrt 3}\right)} \cos^2\varphi\,\textrm d\varphi\\ &= \dfrac{4\sqrt3}{9} \left[ \varphi+ \dfrac12\sin2\varphi\right]_{\operatorname{arctan} \left( {\scriptstyle 1\over\scriptstyle\sqrt 3}\right) }^{\operatorname{arctan} \left({\scriptstyle 3\over\scriptstyle\sqrt 3}\right)}\\ &= \dfrac{4\sqrt3}{9} \left( \operatorname{arctan} \left({\scriptstyle 3\over\scriptstyle\sqrt 3}\right) - \operatorname{arctan} \left({\scriptstyle 1\over\scriptstyle\sqrt 3}\right)\right) \\ &\phantom{=X}+ \dfrac{2\sqrt3}{9} \left( \sin\left( 2\operatorname{arctan} \left({\scriptstyle 3\over\scriptstyle\sqrt 3}\right)\right) - \sin\left( 2\operatorname{arctan} \left({\scriptstyle 1\over\scriptstyle\sqrt 3}\right)\right) \right).\end{aligned}\]

8. \[\dfrac{1}{x-2}-\dfrac{1}{x-1} = \dfrac{1}{(x-2)(x-1)}.\] En élevant au carré, \[\dfrac{1}{(x^2-3x+2)^2}=\dfrac{1}{(x-2)^2}+\dfrac{1}{(x-1)^2} -\dfrac{2}{(x-2)(x-1)} = \dfrac{1}{(x-2)^2}+\dfrac{1}{(x-1)^2} - \dfrac{2}{x-2}+\dfrac{2}{x-1}.\] D’où \[\begin{aligned} \int_0^{-1} \dfrac{\textrm dx}{(x^2-3x+2)^2} &= \left[ -\dfrac{1}{x-2}-\dfrac{1}{x-1}-\ln(2-x) + \ln(1-x) \right]_0^{-1} \\ &= \dfrac13 + \dfrac12 + \ln2 - \dfrac12 -1 + \ln 2\\ &= -\dfrac23 + 2\ln2.\end{aligned}\]

9. Le lecteur vérifiera que \[\int \dfrac{(x-1)\,\textrm dx}{(x^2+1)^2(x+2)}= - {3 \over 25} \log \left( \left|x +2\right|\right) + {3 \over 50} \ln \left(x^2 +1 \right) - {17 \over 50} {\rm arctg} \left(x\right) - \dfrac{1}{10}{x +3 \over x^2 +1} +C\] et que \[\int_{0}^{1} \dfrac{(x-1)\,\textrm dx}{(x^2+1)^2(x+2)}= \dfrac{3}{25}\ln3 +\dfrac{9}{50}\ln2 - \dfrac{17\pi}{200} + \dfrac{1}{10} .\]