求不定积分∫ lnx ⼀ x^1⼀2 dx

2024-12-03 07:21:59
推荐回答(3个)
回答(1):

∫ lnx / x^1/2 dx
=2∫lnxd[x^(1/2)],利用分步积分得到:
=2lnx*x^(1/2)-2∫x^(1/2)dlnx
=2x^(1/2)lnx-2∫x^(1/2)/x dx
=2x^(1/2)lnx-2∫x^(-1/2)dx
=2x^(1/2)lnx-4x^(1/2)+c

回答(2):

1/x^(3/2)-1/2*ln(x)/x^(3/2)

回答(3):

分部积分法 u=lnx v=2x^1/2