Question

求不定积分：∫[f(x)/f′(x)-f²(x)f″(x)/f′(x)³]dx

Answer 1

∫ [ƒ(x)/ƒ'(x) - ƒ(x)²ƒ''(x)/ƒ'(x)³] dx
= ∫ ƒ(x)/ƒ'(x) dx - ∫ ƒ(x)²ƒ''(x)/ƒ'(x)³ dx
= ∫ ƒ(x)/ƒ'(x) dx - ∫ ƒ(x)²/ƒ'(x)³ d[ƒ'(x)]
= ∫ ƒ(x)/ƒ'(x) dx - ∫ ƒ(x)² d[- 1/(2ƒ'(x)²)]
= ∫ ƒ(x)/ƒ'(x) dx + (1/2)∫ ƒ(x)² d[1/ƒ'(x)²]
= ∫ ƒ(x)/ƒ'(x) dx + ƒ(x)²/[2ƒ'(x)²] - (1/2)∫ 1/[ƒ'(x)²] d[ƒ(x)²]，分部积分法
= ∫ ƒ(x)/ƒ'(x) dx + ƒ(x)²/[2ƒ'(x)²] - (1/2)∫ 1/[ƒ'(x)²] * 2ƒ(x) * ƒ'(x) dx
= ∫ ƒ(x)/ƒ'(x) dx + ƒ(x)²/[2ƒ'(x)²] - ∫ ƒ(x)/ƒ'(x) dx，第一和第三项抵消
= ƒ(x)²/[2ƒ'(x)²] + C

Answer 2

∫[ƒ(x)/ƒ'(x) - ƒ(x)²ƒ''(x)/ƒ'(x)³] dx
=∫[f(x)/f′(x)]•(f′(x)²-ƒ(x)²ƒ''(x))/ƒ'(x)²)dx
=∫[f(x)/f′(x)]•(f(x)/ƒ'(x))′dx
=∫[f(x)/f′(x)]d(f(x)/ƒ'(x))
=1/2• ƒ(x)²/ƒ'(x)² + C