Question

求根号下(x/(x+1))的不定积分

Answer 1

∫ √[x/(x + 1)] dx
= ∫ √x/√(x + 1) * √x/√x dx
= ∫ x/√(x² + x) dx
= ∫ x/√[(x + 1/2)² - 1/4] dx
令x + 1/2 = 1/2 * secz，dx = 1/2 * secztanz dz，secz = 2x + 1，tanz = 2√(x² + x)
= ∫ (1/2 * secz - 1/2)/|1/2 * tanz| * (1/2 * secztanz dz)
= (1/2)∫ (secz - 1) * secz dz
= (1/2)∫ sec²z dz - (1/2)∫ secz dz
= (1/2)tanz - (1/2)ln|secz + tanz| + C
= (1/2) * 2√(x² + x) - (1/2)ln|2x + 1 + 2√(x² + x)| + C
= √(x² + x) - (1/2)ln|2x + 1 + 2√(x² + x)| + C

另解：
令u² = x/(x + 1)
x = u²/(1 - u²)
dx = [2u(1 - u²) - u²(- 2u)]/(1 - u²)² du = 2u/(1 - u²)² du
∫ √[x/(x + 1)] dx
= ∫ u * 2u/(1 - u²)² du
= 2∫ u²/(1 - u²)² du，令u = sinz，du = cosz dz，cosz = √(1 - u²)，或用部分分式
= 2∫ sin²z/|cos²z|² * (cosz dz)
= 2∫ secztan²z dz
= 2∫ secz(sec²z - 1) dz
= 2∫ sec³z dz - 2∫ secz dz
= secztanz + ∫ secz dz - 2∫ secz dz
= secztanz - ln|secz + tanz| + C
= 1/√(1 - u²) * u/√(1 - u²) - ln|1/√(1 - u²) + u/√(1 - u²)| + C
= u/(1 - u²) - ln|1 + u| + (1/2)ln|1 - u²| + C
= √[x/(x + 1)] * 1/[1 - x/(x + 1)] - ln|1 + √[x/(x + 1)]| + (1/2)ln|1 - x/(x + 1)| + C
= √x/√(x + 1) * (x + 1) - ln|[√(x + 1) + √x]/√(x + 1)] + (1/2)ln|[(x + 1) - x]/(x + 1)| + C
= √x√(x + 1) - ln|√x + √(x + 1)| + (1/2)ln|x + 1| - (1/2)ln|x + 1| + C
= √(x² + x) - ln|√x + √(x + 1)| + C

两个答案都可以互相转换的。

追问

2∫ sec³z dz - 2∫ secz dz是怎么变成
secztanz - ln|secz + tanz| + C 的？

追答

∫ sec³z dz这个积分嘛，用分部积分法解
I = ∫ sec³z dz = ∫ secz dtanz
= secztanz - ∫ tanz dsecz
= secztanz - ∫ tanz * secztanz dz
= secztanz - ∫ tan²z * secz dz
= secztanz - ∫ (sec²z - 1) * secz dz
= secztanz - I + ∫ secz dz
2I = secztanz + ∫ secz dz
I = (1/2)secztanz + (1/2)∫ secz dz
至于∫ secz dz
= ∫ secz * (secz + tanz)/(secz + tanz) dz
= ∫ (secztanz + sec²z)/(secz + tanz) dz
= ∫ d(secz + tanz)/(secz + tanz)
= ln|secz + tanz| + C，自己看着代入吧