phmg.net
当前位置:首页 >> 积分x2/(1+x4)Dx >>

积分x2/(1+x4)Dx

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) ...

∫ 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...

如图所示:

解:题目中的被积函数是“(x^2-1)/(x^4+1)"吗?若是,分享一种解法。 分子分母同除以x^2,设x+1/x=t。∫(1-1/x^2)dx/(x^2+1/x^2)=∫d(x+1/x)/[(x+1/x)^2-2]=∫dt/(t^2-2)]=(√2/4)ln丨(t-√2)/(t+√2)丨+C。 ∴∫(x^2-1)dx/(x^4+1)=(√2/4)ln丨(x^2+1-√2x...

∫(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

显然被积函数是奇函数 积分限关于原点对称 所以原式=0

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

∫ (1 + x²) / (1 + x^4) dx = ∫ (1 + 1 / x²) / (x² + 1 / x²) dx = ∫ d(x - 1 / x) / [(x - 1 / x)² + 2] = (1 / √2) arctan[(x - 1 / x) / √2] + 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