Question

∫1／﹙x*根号下1-x∧2）要详细过程.

Answer 1

令x = sinz,dx = cosz dz

∫ 1/[x√(1 - x²)] dx

= ∫ 1/(sinz * cosz) * cosz dz

= ∫ cscz dz

= ln(cscz - cotz) + C

= ln[1/sinz - √(csc²z - 1)] + C

= ln[1/sinz - √(1 - sin²z)/sinz] + C

= ln[1/x - √(1 - x²)/x] + C

= ln[1 - √(1 - x²)] - ln(x) + C

对数后面要绝对号,你知道吧