∫ dx/x根号(a2+x2)积分怎么求
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∫ dx/x根号(a2+x2)积分怎么求
![∫ dx/x根号(a2+x2)积分怎么求](/uploads/image/z/13441783-31-3.jpg?t=%E2%88%AB+dx%2Fx%E6%A0%B9%E5%8F%B7%28a2%2Bx2%29%E7%A7%AF%E5%88%86%E6%80%8E%E4%B9%88%E6%B1%82)
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
再问: 若改为x2-a2呢?
再答: 令x = a * secz,dx = a * secztanz 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
再问: 若改为x2-a2呢?
再答: 令x = a * secz,dx = a * secztanz dz