求∫e1sin(1nx)dx的值
求∫e1sin(1nx)dx的值
【正确答案】:I=∫e1sin(lnx)dx=xsin(lnx)∣e1-fe1xcox(lnx)•(1/x)dx=esinl-xcos(lnx)∣e1-fe1sin(lnx)dx=esinl-ecosl+1-I∴I=(e/2)(sinl-cosl)+1/2.
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