求不定积分∫根号下(x^2-a^2) dx

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