在做考研嗎,這種是特殊積分啊
∫√x/lnx dx
=∫x^(1/2)/lnx dx
令d(lnx)=(1/x)dx→dx=xd(lnx)
=∫x^(1/2)/lnx·xd(lnx)
=∫x^(3/2)/lnx d(lnx)
令d[(3/2)lnx]=(3/2)d(lnx)→d(lnx)=d[(3/2)lnx]/(3/2)
=∫x^(3/2)/[(3/2)lnx] d[(3/2)lnx]
=∫e^[lnx^(3/2)]/[lnx^(3/2)] d[lnx^(3/2)]
令u=lnx^(3/2)→du=d[lnx^(3/2)]
=∫e^u/u du
=Ei(u)+C
=Ei[lnx^(3/2)]+C
=Ei[(3/2)lnx)]+C
Ei(x)是指數(shù)積分函數(shù),而[Ei(x)]'=e^x/x
所以∫e^x/x=Ei(x)+C