请问下这道高数题怎么做？不定积分
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∫(1-x)/√(9-x^2)dx
=∫1/√(9-x^2)dx-∫x/√(9-x^2)dx
=∫1/√[1-(x/3)^2]d(x/3)+1/2∫1/√(9-x^2)d(9-x^2)
=arcsin(x/3)+1/2×1/(-1/2+1)(9-x^2)^(-1/2+1)+C
=arcsin(x/3)+(1/4)√(9-x^2)+C
=∫1/√(9-x^2)dx-∫x/√(9-x^2)dx
=∫1/√[1-(x/3)^2]d(x/3)+1/2∫1/√(9-x^2)d(9-x^2)
=arcsin(x/3)+1/2×1/(-1/2+1)(9-x^2)^(-1/2+1)+C
=arcsin(x/3)+(1/4)√(9-x^2)+C
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