第1个回答 推荐于2017-10-18
1
∫arcsinxdx/x^2
=∫arcsinxd(-1/x)
= -arcsinx/x+∫dx/x√(1-x^2)
x=sinu ∫dx/x√(1-x^2)=∫du/sinu=∫-dcosu/(1-cosu)(1+cosu)
=(-1/2)ln|(1+cosu)/(1-cosu)|
=ln|(1-cosu)/sinu|+C
=ln|1/x-√(1-x^2)/x|+C
=-arcsinx/x+ln|1/x-√(1-x^2)/x| +C
2
∫lnxdx/(1-x)^2=∫lnxdx/(x-1)^2=-∫lnxd(1/(x-1))
= -lnx/(x-1)+∫dx/[x(x-1)]
=-lnx/(x-1)+∫[x-(x-1)]dx/[x(x-1)]
=-lnx/(x-1)+ln|x-1|-lnx+C本回答被网友采纳