Question

求不定积分∫根号下(x^2-a^2) dx

Answer 1

答案：(x/2)√(x² - a²) - (a²/2)ln|x + √(x² - a²)| + C

令x = a * secz,dx = a * secztanz dz,假设x > a

∫ √(x² - a²) dx

= ∫ √(a²sec²z - a²) * (a * secztanz dz)

= a²∫ tan²z * secz dz

= a²∫ (sec²z - 1) * secz dz

= a²∫ sec³z dz - a²∫ secz dz

= a²M - a²N

M = ∫ sec³z dz = ∫ secz dtanz

= secztanz - ∫ tanz dsecz

= secztanz - ∫ tanz * (secztanz dz)

= secztanz - ∫ (sec²z - 1) * secz dz

= secztanz - M + N

2M = secztanz + N => N = (1/2)secztanz + N/2

原式= (a²/2)secztanz + a²N/2 - a²N

= (a²/2)secztanz - (a²/2)∫ secz dz

= (a²/2)secztanz - (a²/2)ln|secz + tanz| + C

= (a²/2)(x/a)[√(x² -a²)/a] - (a²/2)ln|x/a + √(x² - a²)/a| + C

= (x/2)√(x² - a²) - (a²/2)ln|x + √(x² - a²)| + C