作业帮(x^6+1)/(x^4+1) 的不定积分
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/16 13:50:53
(x^6+1)/(x^4+1) 的不定积分
(x^6+1)/(x^4+1) 的不定积分
(x^6+1)/(x^4+1) 的不定积分
先将分母因式分解:
x^4+1=(x^2+1)^2-2x^2=(x^2+√2x+1)(x^2-√2x+1)
再将被积函数运用综合除法、待定系数法、凑导数等方法变形：
(x^6+1)/(x^4+1)=(x^6+x^2+1-x^2)/(x^4+1)
=x^2+(1-x^2)/(x^4+1).(综合除法得到)
=x^2+(1/2)[(√2x+1)/(x^2+√2x+1)-(√2x-1)/(x^2-√2x+1)].(待定系数法得到)
=x^2+(1/4)[(2√2x+2)/(x^2+√2x+1)-(2√2x-2)/(x^2-√2x+1)]
=x^2+(√2/4)[(2x+√2)/(x^2+√2x+1)-(2x-√2)/(x^2-√2x+1)]
=x^2+(√2/4)[(x^2+√2x+1)'/(x^2+√2x+1)-(x^2+√2x+1)'/(x^2-√2x+1)].(凑导数得到)
于是原不定积分
∫(x^6+1)/(x^4+1)dx
=∫{x^2+(√2/4)[(x^2+√2x+1)'/(x^2+√2x+1)-(x^2+√2x+1)'/(x^2-√2x+1)]}dx
=∫x^2dx+(√2/4)∫[(x^2+√2x+1)'/(x^2+√2x+1)]dx-(√2/4)∫[(x^2+√2x+1)'/(x^2-√2x+1)]dx
=∫x^2dx+(√2/4)∫[1/(x^2+√2x+1)]d(x^2+√2x+1)-(√2/4)∫[1/(x^2-√2x+1)]d(x^2-√2x+1)
=(1/3)x^3+(√2/4)ln|x^2+√2x+1|-(√2/4)ln|x^2-√2x+1|+c
=(1/3)x^3+(√2/4)ln|(x^2+√2x+1)/(x^2-√2x+1)|+c
=(1/3)x^3+(√2/4)ln[(x^2+√2x+1)/(x^2-√2x+1)]+c (c是任意常数)