求根号下（9—X^2）的不定积分
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∫√(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
=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
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