不定积分怎么求

2024-11-28 18:39:07
推荐回答(1个)
回答1:

1、原式=∫ [2 - 5*(2/3)^x] dx
=2x - 5*(2/3)^xln(2/3) + C

2、∫ (1-x)²/√x dx
=∫ (1-2x+x²)/√x dx
=∫ [x^(-1/2)-2x^(1/2)+x^(3/2)] dx
=2x^(1/2)-(4/3)x^(3/2)+(2/5)x^(5/2)+C

3、∫ cos²(x/2) dx
=(1/2)∫ (1+cosx) dx
=(1/2)x + (1/2)sinx + C

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