不定积分根号(x^2+a^2)^1/2dx

Question

不定积分 根号(x^2+a^2) ^1/2dx

Answer 1

∫ √(x² + a²) dx
x = atanz,dx = asec²z dz
原式 = ∫ (asecz)(asec²z) dz
= a²∫ sec³z dz = a²I
I = ∫ sec³z dz = ∫ secz dtanz
= secztanz - ∫ tanz dsecz
= secztanz - ∫ tan²zsecz dz
= secztanz - ∫ (sec²z - 1)secz dz
= secztanz - I + ∫ secz dz
2I = secztanz + ∫ secz dz
I = (1/2)secztanz + (1/2)ln|secz + tanz| + C,where ∫ secz dz = ln|secz + tanz|
= (1/2)[x/a][√(x² + a²)/a] + (1/2)ln|x/a + √(x² + a²)/a| + C
= [x/(2a²)]√(x² + a²) + (1/2)ln|x + √(x² + a²)| + C
∴原式 = a²I = (x/2)√(x² + a²) + (a²/2)ln|x + √(x² + a²)| + C