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来源：学生作业帮助网 编辑：六六作业网 时间：2025/01/11 20:17:17
======∫(x²-4x+4)/√xdx=∫[x^(3/2)-4x^(1/2)+4x^(-1/2)]dx=2/5x^(5/2)-8/3x^(3/2)-2x^(1/2)+C承认以上复制粘贴另
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∫ (x²-4x+4)/√x dx
=∫ [x^(3/2) -4x^(1/2) +4x^(-1/2)]dx
=2/5 x^(5/2) -8/3x^(3/2) -2x^(1/2) +C
承认以上复制粘贴另一个答案的.然后第三行应改为
=2/5 x^(5/2) -8/3x^(3/2) +2x^(1/2) +C
因为x^n的不定积分公式是（1/(n+1）)*x^(n+1),带入n=-1/2即得.