∫(x−1)e2xdx=12∫(x−1)de2x=12(x−1)e2x−12∫e2xdx
=12(x−1)e2x−14e2x+C
⇒I=[12(x−1)e2x−14e2x]|10=14(3−e2)
Thẻ
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