求不定积分dx/（x*根号下（1-x^2）)

要用换元法

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答：
∫ {1/[x√(1-x^2)]} dx 设x=sint，-π/卜郑2<t<厅弊指π/2
=∫ [1/(sintcost)] d(sint)
=∫ (1/扮配sint) dt
=ln |sect+tant|+C
=ln |1/x+x/√(1-x^2)|+C

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