∫(0~4) (x + 2)/√(2x + 1) dx
令u² = 2x + 1,2udu = 2dx
当x = 0,u = 1,当x = 4,u = 3
= ∫(1~3) [(u² - 1)/2 + 2] * 1/u * udu
= ∫(1~3) (u²/2 + 3/2) du
= (u³/6 + 3u/2)_1^3
= 9 - 5/3
= 22/3
若是：∫(0~4) (x + 2)/(√2x + 1) dx
令x + 2 = A(√2x + 1) + B
x + 2 = √2Ax + A + B
1 = √2A => A = 1/√2
2 = A + B => B = 2 - 1/√2
∫(0~4) (x + 2)/(√2x + 1) dx
= ∫(0~4) [(1/√2)(√2x + 1) + (2 - 1/√2)]/(√2x + 1) dx
= (1/√2)∫(0~4) dx + (2 - 1/√2)∫(0~4) dx/(√2x + 1)
= (1/√2) · x |_0^4 + (2 - 1/√2)(1/√2)∫(0~4) d(√2x + 1)/(√2x + 1)
= (1/√2) · 4 + (√2 - 1/2)ln(√2x + 1)_0^4
= 2√2 + (√2 - 1/2)ln(4√2 + 1) ≈ 4.56你好。 令u² = 2x + 1，2udu = 2dx这步很疑惑阿。 例如： d(√2x+1)=(2x+1)^1/2=1/2（2x+1）^-1/2*dx=1/2*1/√(2x+1)*dx 不是应该前面还要乘以1/2吗？d√(2x + 1) = 1/[2√(2x + 1)] · (2x + 1)' dx <== d/dt √t = 1/(2√t),t = 2x + 1 = 1/[2√(2x + 1)] · (2) dx = dx/√(2x + 1)，2都抵消了