www.whkt.net > 求解一道不定积分1/x^4(1+x^2)Dx

求解一道不定积分1/x^4(1+x^2)Dx

求解一道不定积分1/x^4(1+x^2)dx=∫½[1+cos(2x)]dx =∫½dx+∫½cos(2x)dx =∫½dx+¼

求不定积分1/(x^4(1+x^2))∫1/(x^4(1+x^2))dx=∫(x^2+1-x^2)/(x^4(1+x^2))dx=∫[1/x^4-1/(x

求不定积分1/(x^4(1+x^2))∫1/(x^4(1+x^2))dx =∫(x^2+1-x^2)/(x^4(1+x^2))dx =∫[1/x^4-1/(x^2(1+x^2)]dx =-1

【∫1/x^4(1+x^2)dx怎么求啊?】我按1/[x⁴(1+x²)]来做,如果按(1/x⁴)(1+x²)做太简单了

求不定积分∫1/(x^4*√(1+x^2))dxx=tant,dx=(sect)^2dt原积分=S1/((tant)^4*sect)*(sec)^2dt=Scost^3/sint^4 dt=

一道很麻烦的不定积分∫ 1/(x^4+x^2+1) dx∫1/(x⁴+x²+1) dx = (1/2)∫(x+1)/(x²+x+1) dx - (1/2)∫(x-1)/(x²

的函数的不定积分 :1/(x^4(1+x^2))dx 积分符号没写的,求-1/3X^-3+1/X+arctanX+C

求不定积分∫[(x^4)/(1+x^2)]dx=∫x^4/(1+x²)]dx=∫[(x^4-1)+1]/(1+x²)]dx=∫(x^4-1)/(1+

不定积分1/【x^4乘以√(1-x^2)】dx设x=siny,则dx=cosydy故∫dx/[x^4√(1-x²)]=∫cosydy/[(siny)^4*cosy]=∫

求不定积分(x^4/1+x^2)dx∫(x^4/1+x^2)dx=∫(x^4+x²-x²-1+1)/(1+x^2)dx=∫(x²

友情链接:dzrs.net | wkbx.net | fpbl.net | jinxiaoque.net | famurui.com | 网站地图

All rights reserved Powered by www.whkt.net

copyright ©right 2010-2021。
www.whkt.net内容来自网络,如有侵犯请联系客服。zhit325@qq.com