∫ dx/x根号(a2+x2)积分怎么求
问题描述：
∫ dx/x根号(a2+x2)积分怎么求
答
令x = a * tanz,dx = a * sec²z dz
sinz = x/√(a² + x²),cscz = √(a² + x²)/x,cotz = 1/tanz = a/x
∫ dx/[x√(a² + x²)]
= ∫ 1/[(a * tanz) * |a * secz|] * (a * sec²z dz)
= (1/a)∫ cscz dz
= (1/a)ln|cscz - cotz| + C
= (1/a)ln|√(a² + x²)/x - a/x| + C
= (1/a)ln| [√(a² + x²) - a]/x | + C若改为x2-a2呢？令x = a * secz，dx = a * secztanz dz