大一高数定积分

2025-12-14 09:41:02
推荐回答(2个)
回答1:

春雨(李商隐)

回答2:

∫lnxdx/x
=∫lnxdlnx
=(1/2)(lnx)^2+c

∫sinx(cosx)^3dx
=-∫(cosx)^3dcosx
=-(1/4)cos^4x+c

∫arctgxdx
=xarctgx-∫x/(1+x^2)dx
=xarctgx-(1/2)∫d(x^2+1)/(x^2+1)
=xarctgx-(1/2)ln(x^2+1)+c