Question

∫x^2/√(1-x^2)dx 的不定积分

Answer 1

∫x^2/√(1-x^2)dx=-∫ -2x^2/2√(1-x^2) dx

=-∫ x d√(1-x^2)

=-x√(1-x^2)+∫√(1-x^2)dx

其中,解∫√(1-x^2)dx

令x=sint

dx=costdt

则∫√(1-x^2)dx= ∫(cost)^2 dt

=(1/2)∫(1+cos2t) dt

=(1/2)cost+(1/4)∫cos2t d2t

=(1/2)cost+(1/4)sin2t

=(1/2)√(1-x^2)+(1/2)x√(1-x^2)+C

一加最后是：

(1/2)√(1-x^2)-(1/2)x√(1-x^2)+C