Question

求解：∫（(x平方+a平方）的三次方再开根号）dx

Answer 1

令x = a • tanθ，dx = a • sec²θ dθ
∫ (x² + a²)^(3/2) dx
= ∫ (a² • tan²θ + a²)^(3/2) • (a • sec²θ dθ)
= ∫ a^(2 • 3/2) • (secθ)^(2 • 3/2) • (a • sec²θ dθ)
= ∫ a⁴ • sec⁵θ dθ
= a⁴J
J = ∫ sec⁵θ dθ，用递推公式∫ secⁿx dx = (secⁿ⁻²xtanx)/(n - 1) + [(n - 2)/(n - 1)]∫ secⁿ⁻²x dx得
= (1/4)(sec³θtanθ) + (3/4)∫ sec³θ dθ
= (1/4)(sec³θtanθ) + (3/4)[(1/2)secθtanθ + (1/2)∫ secθ dθ]
= (1/4)(sec³θtanθ) + (3/8)secθtanθ + (3/8)ln|secθ + tanθ| + C
= (1/4)[(x² + a²)^(3/2)/a³][x/a] + (3/8)[√(x² + a²)/a][x/a] + (3/8)ln|x/a + √(x² + a²)/a| + C
= [x/(4a⁴)](x² + a²)^(3/2) + [3x/(8a²)]√(x² + a²) + (3/8)ln|x + √(x² + a²)| + C
原式a⁴J
= (x/4)(x² + a²)^(3/2) + (3a²x/8)√(x² + a²) + (3a⁴/8)ln|x + √(x² + a²)| + C
= (x/8)(5a² + 2x²)√(x² + a²) + (3a⁴/8)ln|x + √(x² + a²)| + C

由于tanθ = x/a，所以secθ = √(tan²θ + 1) = √(x²/a² + 1) = √(x² + a²)/a