令u = 1 + x√x = 1 + x^(3/2)du = (3/2)√x dxx = 1 --> u = 2x = 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)...
