Calculer :
\(\int_{2}^{3} {\scriptstyle 1\over\scriptstyle x\left(x-1\right)}\,\textrm{d}x\)
\(\int_{2}^{3} {\scriptstyle 2x+1\over\scriptstyle x^2-1}\,\textrm{d}x\)
\(\int_{0}^{1} {\scriptstyle x^2-1\over\scriptstyle x^2+4x-5}\,\textrm{d}x\)
\(\int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle x+1\over\scriptstyle\left(x^2+1\right)\left(x-2\right)}\,\textrm{d}x\)
\(\int_{0}^{1} {\scriptstyle 1\over\scriptstyle\left(x^2+2x+5\right)\left(x+2\right)}\,\textrm{d}x\)
\(\int_{0}^{2} {\scriptstyle x\over\scriptstyle\left(x^2+x+1\right)\left(x+1\right)}\,\textrm{d}x\)
\(\int_{0}^{1} {\scriptstyle 1\over\scriptstyle\left(x^2+x+1\right)^2}\,\textrm{d}x\)
\(\int_{0}^{-1} {\scriptstyle 1\over\scriptstyle\left(x^2-3x+2\right)^2}\,\textrm{d}x\)
\(\int_{0}^{1} {\scriptstyle x-1\over\scriptstyle\left(x^2+1\right)^2\left(x+2\right)}\,\textrm{d}x\)
\({\scriptstyle 1\over\scriptstyle x\left(x-1\right)} = {\scriptstyle-1\over\scriptstyle x} + {\scriptstyle 1\over\scriptstyle x-1}\) donc \(\int_{2}^{3} {\scriptstyle 1\over\scriptstyle x\left(x-1\right)}\,\textrm{d}x = \Bigl[ -\ln\left|x\right| + \ln\left|x-1\right| \Bigr]_{2}^{3}=\ln \left({\scriptstyle 4\over\scriptstyle 3}\right)\)
\(\int_{2}^{3} {\scriptstyle 2x+1\over\scriptstyle x^2-1}\,\textrm{d}x = {\scriptstyle 3\over\scriptstyle 2} \int_{2}^{3} {\scriptstyle 1\over\scriptstyle x-1}\,\textrm{d}x + \int_{1}^{2} {\scriptstyle 1\over\scriptstyle x+1}\,\textrm{d}x={\scriptstyle 1\over\scriptstyle 2}\Bigl[ 3\ln \left|x-1\right|+ \ln\left|x+1\right| \Bigr]_{2}^{3}={\scriptstyle 1\over\scriptstyle 2}\left(3\ln 2 + \ln 4 - \ln 3\right)\)
\(\int_{0}^{1} {\scriptstyle x^2-1\over\scriptstyle x^2+4x-5}\,\textrm{d}x = \int_{0}^{1} \left({\scriptstyle x+1\over\scriptstyle x+5}\right)\,\textrm{d}x = \int_{0}^{1} 1-{\scriptstyle 4\over\scriptstyle x+5}\,\textrm{d}x=\Bigl[ x-4\ln \left|x+5\right| \Bigr]_{0}^{1} = 1-4\ln 6 + 4\ln 5\).
\(\int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle x+1\over\scriptstyle\left(x^2+1\right)\left(x-2\right)}\,\textrm{d}x={\scriptstyle 3\over\scriptstyle 5} \int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle 1\over\scriptstyle x-2}\,\textrm{d}x - {\scriptstyle 3\over\scriptstyle 5} \int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle x+{\scriptstyle 1\over\scriptstyle 2}\over\scriptstyle x^2+1}\,\textrm{d}x = \Bigl[ {\scriptstyle 3\over\scriptstyle 5}\ln \left|x-2\right| \Bigr]_{0}^{{\scriptstyle 1\over\scriptstyle 2}} - {\scriptstyle 3\over\scriptstyle 5} \int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle x+{\scriptstyle 1\over\scriptstyle 2}\over\scriptstyle x^2+1}\,\textrm{d}x= {\scriptstyle 3\over\scriptstyle 5}\ln \left({\scriptstyle 3\over\scriptstyle 4}\right) -{\scriptstyle 3\over\scriptstyle 5}\int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle x\over\scriptstyle x^2+1}\,\textrm{d}x - {\scriptstyle 1\over\scriptstyle 5}\int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} {\scriptstyle 1\over\scriptstyle x^2+1}\,\textrm{d}x ={\scriptstyle 3\over\scriptstyle 5}\ln \left({\scriptstyle 3\over\scriptstyle 4}\right) + \Bigl[ -{\scriptstyle 3\over\scriptstyle 10} \ln \left|x^2+1\right| - {\scriptstyle 1\over\scriptstyle 5}\operatorname{arctan} x \Bigr]_{0}^{{\scriptstyle 1\over\scriptstyle 2}}={\scriptstyle 3\over\scriptstyle 5} \ln ({\scriptstyle 3\over\scriptstyle 4}) - {\scriptstyle 3\over\scriptstyle 10} \ln({\scriptstyle 5\over\scriptstyle 4}) - {\scriptstyle 1\over\scriptstyle 5} \operatorname{arctan} ({\scriptstyle 1\over\scriptstyle 2})\)
On écrit la décomposition a priori \(\dfrac{1}{(x+2)(x^2+2x+5)} = \dfrac{A}{x-2}+\dfrac{Bx+C}{x^2+2x+5}\).
On écrit la décomposition a priori \(\dfrac{x}{(x+1)(x^2+x+1)} = \dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+x+1}\).
\[\begin{aligned} \int_0^1 \dfrac{\textrm dx}{(x^2+x+1)^2} &= \int_0^1 \dfrac{\textrm dx}{\left( (x+{\scriptstyle 1\over\scriptstyle 2})^2+{\scriptstyle 3\over\scriptstyle 4}\right)^2 } \\ &= \dfrac{16}{9} \int_0^1 \dfrac{\textrm dx}{\left( ({\scriptstyle 2x+1\over\scriptstyle\sqrt 3})^2+1\right)^2 } \\ &= \dfrac{16}{9} \times \dfrac{\sqrt3}{2}\int_{{\scriptstyle 1\over\scriptstyle\sqrt 3}}^{{\scriptstyle 3\over\scriptstyle\sqrt 3}} \dfrac{\textrm du}{(u^2+1)^2}\end{aligned}\] En posant \(u = \tan\varphi,\,\textrm d\varphi= \dfrac{\textrm du}{u^2+1}\), \[\begin{aligned} \int_0^1 \dfrac{\textrm dx}{(x^2+x+1)^2} &= \dfrac{8\sqrt3}{9} \int_{\operatorname{arctan} \left( {\scriptstyle 1\over\scriptstyle\sqrt 3}\right) }^{\operatorname{arctan} \left({\scriptstyle 3\over\scriptstyle\sqrt 3}\right)} \cos^2\varphi\,\textrm d\varphi\\ &= \dfrac{4\sqrt3}{9} \left[ \varphi+ \dfrac12\sin2\varphi\right]_{\operatorname{arctan} \left( {\scriptstyle 1\over\scriptstyle\sqrt 3}\right) }^{\operatorname{arctan} \left({\scriptstyle 3\over\scriptstyle\sqrt 3}\right)}\\ &= \dfrac{4\sqrt3}{9} \left( \operatorname{arctan} \left({\scriptstyle 3\over\scriptstyle\sqrt 3}\right) - \operatorname{arctan} \left({\scriptstyle 1\over\scriptstyle\sqrt 3}\right)\right) \\ &\phantom{=X}+ \dfrac{2\sqrt3}{9} \left( \sin\left( 2\operatorname{arctan} \left({\scriptstyle 3\over\scriptstyle\sqrt 3}\right)\right) - \sin\left( 2\operatorname{arctan} \left({\scriptstyle 1\over\scriptstyle\sqrt 3}\right)\right) \right).\end{aligned}\]
\[\dfrac{1}{x-2}-\dfrac{1}{x-1} = \dfrac{1}{(x-2)(x-1)}.\] 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 \[\begin{aligned} \int_0^{-1} \dfrac{\textrm dx}{(x^2-3x+2)^2} &= \left[ -\dfrac{1}{x-2}-\dfrac{1}{x-1}-\ln(2-x) + \ln(1-x) \right]_0^{-1} \\ &= \dfrac13 + \dfrac12 + \ln2 - \dfrac12 -1 + \ln 2\\ &= -\dfrac23 + 2\ln2.\end{aligned}\]
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 \[\int_{0}^{1} \dfrac{(x-1)\,\textrm dx}{(x^2+1)^2(x+2)}= \dfrac{3}{25}\ln3 +\dfrac{9}{50}\ln2 - \dfrac{17\pi}{200} + \dfrac{1}{10} .\]
Calculer la primitive (préciser l’intervalle ) \[F=\int \dfrac{x^4}{(x+1)^2(x^2+1) } dx\]
C’est une fraction rationnelle. Après décomposition en éléments simples, on trouve \[F = x - \dfrac{{\scriptstyle 1\over\scriptstyle 2}}{(x+1)} - \dfrac{3}{2} \ln \vert x+1 \vert - \dfrac{1}{4} \ln (x^2+1) + C\] où \(C\) est une constante qui dépend de l’intervalle (\(]-\infty,-1[\) ou \(]-1,+\infty[\)) considéré.