精品偷拍一区二区三区,亚洲精品永久 码,亚洲综合日韩精品欧美国产,亚洲国产日韩a在线亚洲

  • <center id="usuqs"></center>
  • 
    
  • 求不定積分∫(x^2/(1+x^4))dx

    求不定積分∫(x^2/(1+x^4))dx
    數(shù)學人氣:888 ℃時間:2020-07-02 10:53:38
    優(yōu)質解答
    令x=tany
    ∫(x^2/(1+x^4))dx
    =∫(tany^2/(1+tany^4))*(1/(cosy)^2)dy
    =∫(siny)^2/((siny)^4+(cosy)^4) dy
    =∫(1/2)(1-cos2y)/(1-4(siny)^2(cosy)^2) dy
    =(1/2)∫(1-cos2y)/(1-(sin2y)^2) dy
    =(1/2)∫1/(1-(sin2y)^2) dy - (1/2)∫cos2y/(1-(sin2y)^2) dy
    =(1/4)∫(1/(cos2y)^2)d(2y) - (1/4)∫1/((1-sin2y)(1+sin2y)) d(sin2y)
    =(1/4)tan2y - (1/8)∫(1/(1-sin2y) + 1/(1+sin2y))d(sin2y)
    =(1/4)tan2y - (1/8)ln((1+sin2y)/(1-sin2y)) + C
    =(1/4)tan2y - (1/4)ln|(siny+cosy)/(siny-cosy)| + C
    =(1/2)tany/(1-(tany)^2) - (1/4)ln|(tany+1)/(tany-1)| + C
    =(1/2)x/(1-x^2) - (1/4)ln|(x+1)/(x-1)| + C
    我來回答
    類似推薦
    請使用1024x768 IE6.0或更高版本瀏覽器瀏覽本站點,以保證最佳閱讀效果。本頁提供作業(yè)小助手,一起搜作業(yè)以及作業(yè)好幫手最新版!
    版權所有 CopyRight © 2012-2024 作業(yè)小助手 All Rights Reserved. 手機版