phmg.net
当前位置:首页 >> ∫Dx/x4(x2+1)=? >>

∫Dx/x4(x2+1)=?

∫ 1/[x^4(x²+1)] dx =∫ (1+x²-x²)/[x^4(x²+1)] dx =∫ (1+x²)/[x^4(x²+1)] dx - ∫ x²/[x^4(x²+1)] dx =∫ 1/x^4 dx - ∫ 1/[x²(x²+1)] dx =-1/(3x³) - ∫ 1/x² dx + ∫ 1/(x²+1) d...

∫(x2+1)/(x4+1)dx=∫(1+1/x2)/(x2+1/x2)dx=∫1/[(x-(1/x))^2+2]d(x-(1/x))=[(√2)/2]arctan{[x-(1/x)]/√2}+C

如图所示:

本题解法技巧较高 ∫ x²/(1+x^4) dx =(1/2)∫ (x²-1+x²+1)/(1+x^4) dx =(1/2)∫ (x²-1)/(1+x^4) dx + (1/2)∫ (x²+1)/(1+x^4) dx 分子分同除以x² =(1/2)∫ (1-1/x²)/(1/x²+x²) dx + (1/2)∫ (1+1/x²...

原式=1/(x^2-1)(x^2+1)=1/2*(1/(x^-1)-1/(x^2+1))=1/4(1/(1-x)-1/(1+x))-1/2*(1/(x^2+1)) 对上式积分得1/4*(ln(x-1)-ln(x+1))+1/2*acrtanx

∫(x2+1)/(x4+1)dx=∫(1+1/x^2)/(x^2+1/x^2)dx=∫{1/[2+(x-1/x)^2]}d(x-1/x) =(√2/2)arctan[(x-1/x)/√2]+c(希望能帮到你,麻烦在我回答的下面点击 “好评”,谢谢你啦^_^)

就是这样化出来的

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) ...

约定:∫[a,b]表示[a,b]上的定积分 要证 1≤∫[0,1](√(x^4+1))dx≤6/5 只需证0≤∫[0,1]((√(x^4+1))-1)dx≤1/5 ∫[0,1]((√(x^4+1))-1)dx≥∫[0,1]((√(0^4+1))-1)dx=0 得∫[0,1]((√(x^4+1))-1)dx≥0 ∫[0,1]((√(x^4+1))-1)dx=∫[0,1](x^4/((√(x^4+1))+1))dx (分...

解: 令x²=u ∫[x/(1+x⁴)]dx =½∫[1/(1+x⁴)]d(x²) =½∫[1/(1+u²)]du =½arctanu +C =½arctan(x²) +C

网站首页 | 网站地图
All rights reserved Powered by www.phmg.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com