求不定积分 ∫xln^2 xdx

2024-11-23 22:45:21
推荐回答(1个)
回答1:

用分部积分法:
∫x(lnx)^2dx
=(1/2)x^2(lnx)^2-∫(1/2)x^2*(1/x)dx
=(1/2)x^2(lnx)^2-∫(1/2)xdx
=(1/2)x^2(lnx)^2-(1/4)x^2+C