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Sagot :
Split the integrand into partial fractions.
[tex]\dfrac{9x+2}{x^2+x-6} = \dfrac{9x+2}{(x-2)(x+3)} = \dfrac a{x-2} + \dfrac b{x+3}[/tex]
[tex]\implies 9x+2 = a(x+3) + b(x-2) = (a+b)x + (3a-2b)[/tex]
[tex]\implies \begin{cases}a+b=9 \\ 3a-2b=2\end{cases} \implies a=4,b=5[/tex]
Then we have
[tex]\displaystyle \int \frac{9x+2}{x^2+x-6} \, dx = 4 \int \frac{dx}{x-2} + 5 \int \frac{dx}{x+3} \\\\ = \boxed{4\ln|x-2| + 5\ln|x+3| + C}[/tex]
which follows from the result
[tex]\displaystyle \int \frac{dx}x = \ln|x|+C[/tex]
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