问题描述： 求根号下（9—X^2）的不定积分 1个回答 分类：数学 2014-12-16 问题解答： 我来补答 ∫√(9-x^2)dx=x√(9-x^2)-∫xd(√9-x^2)=x√(9-x^2)+∫x^2/√(9-x^2)dx=x√(9-x^2)+∫(9-(9-x^2))/√(9-x^2)dx=x√(9-x^2)+∫9/√(9-x^2)dx-∫√(9-x^2)dx从而2∫√(9-x^2)dx=x√(9-x^2)+∫9/√(9-x^2)dx所以∫√(9-x^2)dx=(x√(9-x^2)+9∫d(x/3)/√(1-(x/3)^2))/2=(x√(9-x^2)+9arcsin(x/3))/2 展开全文阅读
