Question

求好心人解一道不定积分∫x^4√(1+x^2)dx

Answer 1

x⁴ = [(x² +1) - 1]² = (x² + 1)² - 2(x² + 1) + 1
∫ x⁴√(1 + x²) dx
= ∫ [(x² + 1)² - 2(x² + 1) + 1]√(1 + x²) dx
= ∫ (x² + 1)^(5/2) dx - 2∫ (x² + 1)^(3/2) dx + ∫ √(x² + 1) dx
令x = tanz，dx = sec²z dz，x² + 1 = sec²z
= ∫ sec⁷z dz - 2∫ sec⁵z dz + ∫ sec³z dz
用递推公式：∫ secⁿx dx = (secⁿ⁻²xtanx)/(n-1) + (n-2)/(n-1) × ∫ secⁿ⁻²x dx
= (1/6)sec⁵ztanz + (5/6)∫ sec⁵z dz - 2∫ sec⁵z dz + ∫ sec³z dz
= (1/6)sec⁵ztanz - (7/6)∫ sec⁵z dz + ∫ sec³z dz
= (1/6)sec⁵ztanz - (7/6)[(1/4)sec³ztanz + (3/4)∫ sec³z dz] + ∫ sec³z dz
= (1/6)sec⁵ztanz - (7/24)sec³ztanz + (1/8)∫ sec³z dz
= (1/6)sec⁵ztanz - (7/24)sec³ztanz + (1/8)[(1/2)secztanz + (1/2)∫ secz dz]
= (1/6)sec⁵ztanz - (7/24)sec³ztanz + (1/16)secztanz + (1/16)ln|secz + tanz| + C
= (1/6)x(1 + x²)^(5/2) - (7/24)x(1 + x²)^(3/2) + (1/16)x√(1 + x²) + (1/16)ln|x + √(1 + x²)| + C
= (x/48)(8x⁴ + 2x² - 3)√(1 + x²) + (1/16)ln|x + √(1 + x²)| + C

Answer 2

好难