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Meilleure réponse !
2014-06-10T23:54:35+02:00
Bonsoir,

\int(\dfrac{1}{x^2-4}+\dfrac{1}{x^2+4})dx=\int\dfrac{1}{x^2-4}dx+\int\dfrac{1}{x^2+4})dx

Or,  \dfrac{1}{x^2-4}=\dfrac{1}{(x-2)(x+2)}\\\\\dfrac{1}{x^2-4}=\dfrac{a}{x-2}+\dfrac{b}{x+2}\\\\\dfrac{1}{x^2-4}=\dfrac{a(x+2)+b(x-2)}{x+2}\\\\1=a(x+2)+b(x-2)\\\\Si\ x=2,\ alors\  1=a(2+2)+b(2-2)\\\\1=4a\\\\a=\dfrac{1}{4}

Si\ x=-2,\ alors\  1=a(-2+2)+b(-2-2)\\\\1=-4b\\\\b=-\dfrac{1}{4}

D'où  \dfrac{1}{x^2-4}=\dfrac{1}{4(x-2)}-\dfrac{1}{4(x+2)}\\\\\int\dfrac{1}{x^2-4}dx=\int(\dfrac{1}{4(x-2)}-\dfrac{1}{4(x+2)})dx\\\\\int\dfrac{1}{x^2-4}dx=\int(\dfrac{1}{4(x-2)}dx-\int\dfrac{1}{4(x+2)})dx\\\\\int\dfrac{1}{x^2-4}dx=\dfrac{1}{4}\int\dfrac{1}{x-2}dx- \dfrac{1}{4}\int\dfrac{1}{x+2}dx\\\\\int\dfrac{1}{x^2-4}dx=\dfrac{1}{4}\ln|x-2|- \dfrac{1}{4}\ln|x+2|+C_1\\\\\int\dfrac{1}{x^2-4}dx=\dfrac{1}{4}(\ln|x-2|- \ln|x+2|)+C_1

\boxed{\int\dfrac{1}{x^2-4}dx=\dfrac{1}{4}\ln|\dfrac{x-2}{x+2}|+C_1}

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\int\dfrac{1}{x^2+4}dx=\int\dfrac{1}{4(\dfrac{x^2}{4}+1)}dx\\\\\int\dfrac{1}{x^2+4}dx=\int\dfrac{1}{4[(\dfrac{x}{2})^2+1]}dx\\\\\int\dfrac{1}{x^2+4}dx=\dfrac{1}{2}\int\dfrac{\frac{1}{2}}{(\dfrac{x}{2})^2+1}dx\\\\\int\dfrac{1}{x^2+4}dx=\dfrac{1}{2}\int\dfrac{\frac{1}{2}}{1+(\dfrac{x}{2})^2}dx\\\\\boxed{\int\dfrac{1}{x^2+4}dx=\dfrac{1}{2}\tan^{-1}(\dfrac{x}{2})+C_2}

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Par conséquent,

\boxed{\int(\dfrac{1}{x^2-4}+\dfrac{1}{x^2+4})dx=\dfrac{1}{4}\ln|\dfrac{x-2}{x+2}|+\dfrac{1}{2}\tan^{-1}(\dfrac{x}{2})+C}