∫(1+x^2)^3/2dx=?没多少积分了,请各位大侠多包涵.∫(1+x^2)^（-3/2）dx=?呢

Question

∫(1+x^2)^3/2dx=?没多少积分了,请各位大侠多包涵.
∫(1+x^2)^（-3/2）dx=?呢

Answer 1

令x = tany,dx = sec²y dy
J = ∫ (1 + x²)^(3/2) dx
= ∫ (1 + tan²y)^(3/2) (sec²y dy)
= ∫ (sec²y)^(3/2) * sec²y dy,恒等式1 + tan²x = sec²x
= ∫ [secy]^5 dy ...(★)
= ∫ sec³y d(tany)
= sec³y tany - ∫ tany d(sec³y)
= sec³y tany - ∫ tany 3sec²y secy tany dy
= sec³y tany - 3∫ tan²y sec³y dy
= sec³y tany - 3∫ (sec²y - 1)sec³y dy
=> J = sec³y tany - 3J + 3K
K = ∫ sec³y dy = ∫ secy d(tany)
= secy tany - ∫ tany d(secy)
= secy tany - ∫ tany secy tany dy
= secy tany - ∫ (sec²y - 1)secy dy
= secy tany - K + ∫ secy dy
2K = secy tany + ∫ secy dy
K = (1/2)secy tany + (1/2)ln|secy + tany|
∴J = sec³y tany - 3J + 3K
4J = sec³y tany + (3/2)secy tany + (3/2)ln|secy + tany|
J = (1/4)sec³y tany + (3/8)secy tany + (3/8)ln|secy + tany| + C
= (1/4)x(1 + x²)^(3/2) + (3/8)x√(1 + x²) + (3/8)ln|x + √(1 + x²)| + C
不想做得这么复杂的话,在(★)处可用降幂公式：
∫ [secy]^n dy = [siny (secy)^(n - 1)]/(n - 1) + [(n - 2)/(n - 1)]∫ [secy]^(n - 2) dy,代入n = 5和n = 3后就计算到了
case 2 简单得多.
∫ (1 + x²)^(-3/2) dx,x = tany,dx = sec²y dy
= ∫ (1 + tan²y)^(-3/2) sec²y dy
= ∫ (sec²y)^(-3/2) sec²y dy
= ∫ 1/sec³y * sec²y dy
= ∫ cosy dy
= siny + C
= x/√(1 + x²) + C