定积分换元法求上限e^2,下限1,1/[x√(1+lnx)]dx的定积分

Answer 1

先求不定积分
∫ lnx/√x dx
=2∫ lnx d(√x) （分部积分法）
=2√xlnx - 2∫ √x/x dx
=2√xlnx - 2∫ 1/√x dx
=2√xlnx - 4√x + C
再把上下限代入相减即可,这个很简单,因为不好输入,我就不帮你写了.
满意请采纳哦,谢谢~

Answer 2

is so heavy but that by prayer and repent