Tính nguyên hàm: \(I=\int \dfrac{\mathrm{dx}}{\mathrm{x}^3-3 \mathrm{x}+2}\)
Tính nguyên hàm: \(I=\int \dfrac{\mathrm{dx}}{\mathrm{x}^3-3 \mathrm{x}+2}\)
Đáp án đúng là: B
Quảng cáo
Ta có: \(I=\int \dfrac{\mathrm{dx}}{(\mathrm{x}-1)^2(\mathrm{x}+2)}=\dfrac{1}{3} \int \dfrac{(\mathrm{x}+2)-(\mathrm{x}-1)}{(\mathrm{x}-1)^2(\mathrm{x}+2)} \mathrm{dx}\)
\(=\dfrac{1}{3} \int \dfrac{\mathrm{dx}}{(\mathrm{x}-1)^2}-\dfrac{1}{3} \int \dfrac{\mathrm{dx}}{(\mathrm{x}-1)(\mathrm{x}+2)}=\dfrac{-1}{3(\mathrm{x}-1)}-\dfrac{1}{9} \int\left(\dfrac{1}{\mathrm{x}-1}-\dfrac{1}{\mathrm{x}+2}\right) \mathrm{dx}\)
\(=-\dfrac{1}{3(\mathrm{x}-1)}-\dfrac{1}{9} \ln \left|\dfrac{\mathrm{x}-1}{\mathrm{x}+2}\right|+C\)
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