$\int(x-1)e^{2x}dx=\frac{1}{2}\int(x-1)de^{2x}=\frac{1}{2}(x-1)e^{2x}-\frac{1}{2}\int e^{2x}dx$
$=\frac{1}{2}(x-1)e^{2x}-\frac{1}{4}e^{2x}+C$
$\Rightarrow I=[\frac{1}{2}(x-1)e^{2x}-\frac{1}{4}e^{2x}]|^1_0=\frac{1}{4}(3-e^2)$
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