Réponse :
Bonjour,
[tex]\int{2(x^3-2x)(3x^2-2)} \, dx =(x^3-2x)^2+C \\\\\int{sin^2x\ cos x} \, dx =\dfrac{sin^3 x}{3}+C \\\\\int\limits^1_0 {\dfrac{ -4e^{5x}*e^{-6x}}{e^{-2x}}} \, dx \\=-4\int\limits^1_0 {e^{x}} \, dx\\\\=-4[e^x]^1_0\\\\=4(1-e)\\\\\\\int\limits^1_{-1} {\dfrac{e^{-x}}{(1+e^{-x})^2} } \, dx \\\\=-\int\limits^1_{-1} {\dfrac{-e^{-x}}{(1+e^{-x})^2} } \, dx \\\\=[\dfrac{1}{1+e^{-x}} ]^1_{-1}\\\\=\dfrac{e}{e+1} -\dfrac{1}{e+1} \\\\==\dfrac{e-1}{e+1}[/tex]
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