要加絕對(duì)值,定義域不同的
∫ √(x² - 1)/x⁴ dx
令x = secz、dx = secztanz dz、cosz = 1/x、sinz = √(x² - 1)/x
當(dāng)x > 1、0 ≤ z < π/2
∫ √(x² - 1)/x⁴ dx
= ∫ tanz/sec⁴z • secztanz dz
= ∫ tan²z/sec³z dz
= ∫ (sec²z - 1)/sec³z dz
= ∫ cosz dz - ∫ cos³z dz
= sinz - ∫ (1 - sin²z) d(sinz)
= sinz - sinz + (1/3)sin³z + C
= (1/3)[ √(x² - 1)/x ]³ + C
= [(x² - 1)^(3/2)]/(3x³) + C
當(dāng)x < - 1、π/2 < z ≤ π
∫ √(x² - 1)/x⁴ dx
= ∫ (- tanz)/sec⁴z • secztanz dz
= - [(x² - 1)^(3/2)]/(3x³) + C
∴合并為[(x² - 1)^(3/2)]/(3|x|³) + C
運(yùn)用倒代換：
∫ √(x² - 1)/x⁴ dx、x = 1/t、dx = - 1/t² dt
= ∫ t⁴√(1/t² - 1) • (- 1/t²) dt
= - ∫ t²√(1 - t²)/t dt
= (1/2)∫ √(1 - t²) d(1 - t²)
= (1/2)(3/2)(1 - t²)^(3/2) + C
= (1/3)(1 - 1/x²)^(3/2) + C
= (x² - 1)^(3/2)/(3|x|³) + C