Nếu \(\int_{2}^{3} \frac{x+2}{2 x^{2}-3 x+1} \mathrm{d} x=a \ln 5+b \ln 3+3 \ln 2(a, b \in \mathbb{Q})\) thì giá trị của P=2a-b là:
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Lời giải:
Báo saiTa có:
\(\begin{array}{l} \int_{2}^{3} \frac{x+2}{2 x^{2}-3 x+1} \mathrm{d} x=\frac{1}{4} \int_{2}^{3} \frac{4 x-3}{2 x^{2}-3 x+1} \mathrm{d} x+\frac{11}{4} \int_{2}^{3} \frac{1}{2 x^{2}-3 x+1} \mathrm{d} x \\ =\frac{1}{4} \int_{2}^{3} \frac{1}{2 x^{2}-3 x+1} \mathrm{d}\left(2 x^{2}-3 x+1\right)+\frac{11}{4} \int_{2}^{3} \frac{1}{(x-1)(2 x-1)} \mathrm{d} x \end{array}\)
\(\begin{array}{l} =\left.\frac{1}{4} \ln \left|2 x^{2}-3 x+1\right|\right|_{2} ^{3}+\frac{11}{4} \int_{2}^{3}\left(\frac{1}{x-1}-\frac{2}{2 x-1}\right) \mathrm{d} x \\ =\left.\frac{1}{4} \ln \left|2 x^{2}-3 x+1\right|\right|_{2} ^{3}+\left.\frac{11}{4} \ln \left|\frac{x-1}{2 x-1}\right|\right|_{2} ^{3}=\frac{1}{4}(\ln 10-\ln 3)+\frac{11}{4}\left(\ln \frac{2}{5}-\ln \frac{1}{3}\right) \end{array}\)
\(=\frac{1}{4} \ln \frac{10}{3}+\frac{11}{4} \ln \frac{6}{5}=\frac{1}{4}(\ln 5+\ln 2-\ln 3)+\frac{11}{4}(\ln 2+\ln 3-\ln 5)\)
\(=-\frac{5}{2} \ln 5+\frac{5}{2} \ln 3+3 \ln 2\)
Khi đó \(a=-\frac{5}{2}, b=\frac{5}{2}\Rightarrow P=-\frac{15}{2}\)