Question

求∫√x／1+x√xdx在上下限1到4的定积分。

Answer 1

令t=√x，则dx＝tdt,积分限变为1到2，原式化为∫ t2/(1+t3)dt=∫ {1/[3(1+t)]+(2t-1)/[3(t2-t+1)]}dt=
（1/3）ln(1+t)+∫ 1/[3(t2-t+1)]d(t2-t+1)，原式化为（1/3）ln(1+t）（1到2）加上
（1/3）ln(t2-t+1) (1到2) 结果是（1/3）ln(3/2)+(1/3)ln3=(1/3)ln(9/2)

Answer 2

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Answer 3

令u = 1 + x√x = 1 + x^(3/2)
du = (3/2)√x dx
x = 1 --> u = 2
x = 4 --> u = 9
∫(1，4) √x/(1 + x√x) dx
= ∫(2，9) √x/u * (2/3)(1/√x) du
= (2/3)∫(2，9) du/u
= (2/3)ln| u | (2，9)
= (2/3)[ln(9) - ln(2)]
= (2/3)ln(9/2)