求不定积分:[(x^2-1)*根号下(1-x^2)

求不定积分:[(x^2-1)*根号下(1-x^2)-3x]/x*根号下(1-x^2)

1个回答

原式=∫[(x^2-1)/x-3/根号下(1-x^2)]dx
=∫[x-(1/x)-3/根号下(1-x^2)]dx
=∫xdx-∫(1/x)dx-3∫[1/根号下(1-x^2)]dx
=(x^2)/2-lnx-3arcsinx+C