nguyên hàm của (2*x^2-8*x+10) /(x^3 +x^2 -4*x-4)

$\displaystyle\int \dfrac{2x^2-8x+10}{x^3+x^2-4x-4} \, dx\\ =\displaystyle\int \dfrac{\dfrac{2}{3}(3x^2+2x-4)-\dfrac{28}{3}x-\dfrac{28}{3}+22}{x^3+x^2-4x-4} \, dx \\ =\dfrac{2}{3}\displaystyle\int \dfrac{3x^2+2x-4}{x^3+x^2-4x-4} \, dx -\dfrac{28}{3}\displaystyle\int \dfrac{x+1}{x^3+x^2-4x-4} \, dx+22\displaystyle\int \dfrac{1}{x^3+x^2-4x-4} \, dx \\ =\dfrac{2}{3}\displaystyle\int \dfrac{d(x^3+x^2-4x-4)}{x^3+x^2-4x-4} \, dx -\dfrac{28}{3}\displaystyle\int \dfrac{x+1}{(x+1)(x-2)(x+2)} \, dx+22\displaystyle\int \dfrac{1}{(x+1)(x-2)(x+2)} \, dx \\ =\dfrac{2}{3}\ln{(|x^3+x^2-4x-4|)}-\dfrac{28}{3}\displaystyle\int \dfrac{1}{(x-2)(x+2)} \, dx+22\displaystyle\int \dfrac{1}{(x+1)(x-2)(x+2)} \, dx +C\\ =\dfrac{2}{3}\ln{(|(x+1)(x-2)(x+2)|)}-\dfrac{28}{3}.\dfrac{1}{4}.\displaystyle\int \left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right) \, dx+22\displaystyle\int\left(\dfrac{1}{12(x-2)}+\dfrac{1}{4(x+2)}-\dfrac{1}{3(x+1)}\right) \, dx +C\\ =\dfrac{2}{3}\ln{(|x+1|)}+\dfrac{2}{3}\ln{(|x+2|)}+\dfrac{2}{3}\ln{(|x-2|)}-\dfrac{7}{3}\ln{(|x-2|)}+\dfrac{7}{3} \ln{(|x+2|)}+\dfrac{11}{6}\ln{(|x-2|)}+\dfrac{11}{2}\ln{(|x+2|)}-\dfrac{22}{3}\ln{(|x+1|)}+C\\ =\dfrac{-20}{3}\ln{(|x+1|)}+\dfrac{17}{2}\ln{(|x+2|)}+\dfrac{1}{6}\ln{(|x-2|)}+C$

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