题目

求不定积分∫r^3 √(1+r^2) dr,

提问时间：2021-03-03

答案

令r = tanθ,dr = sec²θdθ
√(1 + r²) = √(1 + tan²θ) = √sec²θ = secθ
∫ r³√(1 + r²) dr
= ∫ (tan³θsecθ)(sec²θ) dθ
= ∫ tan³θsec³θ dθ
= ∫ tan²θsec²θ d(secθ)
= ∫ (sec²θ - 1)sec²θ d(secθ)
= ∫ (sec⁴θ - sec²θ) d(secθ)
= (1/5)sec⁵θ - (1/3)sec³θ + C
= (1/5)(1 + r²)^(5/2) - (1/3)(1 + r²)^(3/2) + C,进一步因式分解.
= (1/15)(3r² - 2)(1 + r²)^(3/2) + C