Question

求x*2+1/x*4-x*2+1的不定积分怎么写？ 还有e*3x+e*x/e*4x-e*2x+1的

Answer 1

设(x^2+1)/(x^4-x^2+1)=（ax+b）/(x^2-√3x+1)+(cx+d)/(x^2+√3x+1),
去分母得x^2+1=(ax+b)(x^2+√3x+1)+(cx+d)(x^2-√3x+1)
=(a+c)x^3+(√3a+b-√3c+d)x^2+(a+√3b+c-√3d)x+b+d,
比较系数得a+c=0,√3a+b-√3c+d=1,a+√3b+c-√3d=0,b+d=1,
解得a=-√3/6，c=√3/6，b=d=1/2,
所以∫(x^2+1)dx/(x^4-x^2+1)
=(√3/6)∫[(-x+√3）/(x^2-√3x+1）+(x+√3)/(x^2+√3x+1)]dx
=(√3/12){ln[(x^2+√3x+1)/(x^2-√3x+1)]+2√3[arctan(2x-√3)+arctan(2x+√3)]}+c.
设u=e^x,则du=e^x*dx,
∫[e^x+e^(3x)]dx/[e^(4x)-e^(2x)+1]
=∫(1+u^2)du/(u^4-u^2+1),
剩下部分留给您练习，可以吗？

Answer 2