∫√(25-x^2)dx/x的不定积分怎么求
<p>问题：∫√(25-x^2)dx/x的不定积分怎么求<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">高忠江的回答：<div class="content-b">网友采纳 令x=5sinθ,dx=5cosθdθ √(25-x²)=√(25-25sin²θ)=5cosθ ∫√(25-x²)/xdx =∫5cosθ/5sinθ•5cosθdθ =5∫cos²θ/sinθdθ =5∫(1-sin²θ)/sinθdθ =5∫(cscθ-sinθ)dθ =5ln|cscθ-cotθ|+5cosθ+C =5ln|5/x-√(25-x²)/x|+√(25-x²)+C =5ln|5-√(25-x²)|-5ln|x|+√(25-x²)+C
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