Question

求不定积分∫x^2/(x^4-x^2+1)dx ，数学高手们，谢谢了

求不定积分∫x^2/(x^4-x^2+1)dx ，数学高手们，谢谢了

Answer 1

∫dx/(x^2-1+1/x^2)
=∫dx/[(x+1/x)^2-3]
(x+1/x)^2-3这个最大值是正无穷
此时积分最小为0
最小当x=1/x时
此时x=1
积分为∫dx=x=1
所以函数是∫(0,+无穷)a^x(0<a<1)的值
=1/2

Answer 2

∫x^2/(x^4-x^2+1)dx
=1/2∫[x/(x^2-x+1)-x/(x^2-x+1)]dx
∫x/(x^2-x+1)dx=∫1/2 * 1/(x^2-x+1)d(x^2-x+1) + 1/(x^2-x+1)dx
=ln│x^2-x+1│+（2/√3）arctan((2x-1)/√3) +C
2∫x^2/(x^4-x^2+1)dx
= ln│x^2-x+1│+（2/√3）arctan((2x-1)/√3) -(ln│x^2+x+1│+（2/√3）arctan((2x+1)/√3))
=ln│(x^2-x+1)/(x^2+x+1)│+（2/√3）(arctan((2x-1)/√3)-arctan((2x+1)/√3);
∫x^2/(x^4-x^2+1)dx
=1/2* [ ln│(x^2-x+1)/(x^2+x+1)│+（2/√3）(arctan((2x-1)/√3)-arctan((2x+1)/√3) ] +C

Answer 3

∫x^2dx/(x^4-x^2+1)
=(1/2)∫xdx^2/[(x^2+1)^2-3x^2]
=(1/4√3)∫(2√3x)dx^2/[(x^2+1+√3x)(x^2+1-√3x)]
=(1/4√3)[∫dx^2/(x^2+1-√3x)-∫dx^2/(x^2+1+√3x)]
=(1/4√3) [∫d(x^2-√3x+1)/(x^2+1-√3x) +∫√3dx/(x^2-√3x+1)
-∫d(x^2+√3x+1)/(x^2+1+√3x)+∫√3dx/(x^2+√3x+1)]
=(1/4√3)[ln|x^2+1-√3x|/|x^2+1+√3x| +√3∫dx/[(x-√3/2)^2+1/4]+√3∫dx/[(x+√3/2)^2+1/4]
=(1/4√3[ln|x^2+1-√3x|/|x^2+1+√3x| +2√3aratn(2x-√3)+2√3arctan(2x+√3)]+C
=(1/4√3)ln[|x^2+1-√3x|/|x^2+1+√3x|] +(1/2)arctan(2x-√3)+(1/2)arctan(2x+√3)+C