Corrigé du 91 P. 385

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a. For all $x$ in $]1;+\infty[$: \begin{align*} a + \frac b {x-1} + \frac c {(x+1)^2} &=\frac{a(x-1)^2 + b(x-1) + c}{(x-1)^2}& \\ &=\frac{ax^2 -2ax + a + bx - b + c}{(x-1)^2}& \\ &=\frac{ax^2 + (-2a+b)x + c}{(x-1)^2}.& \end{align*} Therefore \[\frac{ax^2 + (-2a+b)x + c}{(x-1)^2} = \frac{x^2 -x + 2}{(x-1)^2}\] if and only if: \[\begin{cases}a=1\\-2a+b = -1\\c = 2\end{cases} \iff \begin{cases}a=1\\b=1\\c=2\end{cases}.\]

b. \begin{align*} I &= \int_2^4 f(x)\mathrm dx& \\ &=\int_2^4\left(1 + \frac 1 {x-1} + \frac 2{(x-1)^2}\right)\mathrm dx& \\ &=\left[x + \ln(x-1) -\frac 2 {x-1}\right]_2^4& \\ &=4 + \ln(4-1) - \frac 2 {4-1} - 2 - \ln(2-1) + \frac 2 {2-1}& \\ &=4 + \ln(3) -\frac 2 3 -\ln(3) + 2& \\ &=6 - \frac 2 3& \\ &=\frac{16}3.& \end{align*}

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