∫arcsin根号(x/1+x)dx

网友回答

分步积分得∫arcsin{[x/(1+x)]^(1/2)}dx
=xarcsin{[x/(1+x)]^(1/2)}
-∫x/2[1-x/(x+1)]^(1/2)*[(x+1)/x]^(1/2)*dx/(x+1)^2
=xarcsin{[x/(x+1)]^(1/2)}-∫x^(1/2)/2(x+1) dx
=xarcsin{[x/(x+1)]^(1/2)}-∫t/2(t^2+1)*2tdt 设x=t^2
=xarcsin{[x/(x+1)]^(1/2)}-∫[1-1/(t^2+1)]dt
=xarcsin{[x/(x+1)]^(1/2)}-t+arctant+C
arctant=arcsin{[x/(x+1)]^(1/2)}
=(x+1)arcsin{[x/(x+1)]^(1/2)}-x^(1/2)+C