Question

高数题这题咋做？

谢谢

Answer 1

这是一个瑕积分，瑕点为x=0
原式=lim[c→0+]∫[c,1]lnx /√x dx
=lim[c→0+]2∫[c,1]lnx d(√x)
=lim[c→0+]2√x lnx|[c,1] - lim[c→0+]2∫[c,1]dx/√x
=0 - lim[c→0+]2√x lnx - lim[c→0+]4√x |[c,1]
=- lim[c→0+]2√x lnx - 4 + lim[c→0+]4√x
=-4+lim[c→0+]2√x (2-lnx)
=-4+lim[t→+∞]2(2+lnt)/√t 【令t=1/x】
=-4+lim[t→+∞]4/√t【∞/∞型，洛必达之后化简】
=-4

Answer 2

∫(0->1) lnx/√x dx
=2∫(0->1) lnx d√x
=2[√x.lnx]|(0->1) - 2∫(0->1) dx/√x
=0 - 4[√x]|(0->1)
=-4