作业帮高数 不定积分 8
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/20 22:04:55
高数 不定积分 8
答：
x/(1+x³)
=1/3×((x+1)/(x²-x+1)-1/(x+1))
=1/3×(1/2×(2x-1)/(x²-x+1)+3/2×1/((x-1/2)²+3/4)-1/(x+1))
=1/6×(2x-1)/(x²-x+1)+1/2×1/((x-1/2)²+3/4)-1/3×1/(x+1)
=1/6×(2x-1)/(x²-x+1)+1/2×4/3×1/((2/√3)×(x-1/2))²+1)-1/3×1/(x+1)
=1/6×(2x-1)/(x²-x+1)+1/√3×(2/√3)/((2x/√3-1/√3))²+1)-1/3×1/(x+1)
所以原积分
=∫[1/6×(2x-1)/(x²-x+1)+1/√3×(2/√3)/((2x/√3-1/√3))²+1)-1/3×1/(x+1)] dx
=(ln|x²-x+1|)/6 - (ln|x+1|)/3 + arctan((2x-1)/√3)/√3 + C
x/(1+x³)
=1/3×((x+1)/(x²-x+1)-1/(x+1))
=1/3×(1/2×(2x-1)/(x²-x+1)+3/2×1/((x-1/2)²+3/4)-1/(x+1))
=1/6×(2x-1)/(x²-x+1)+1/2×1/((x-1/2)²+3/4)-1/3×1/(x+1)
=1/6×(2x-1)/(x²-x+1)+1/2×4/3×1/((2/√3)×(x-1/2))²+1)-1/3×1/(x+1)
=1/6×(2x-1)/(x²-x+1)+1/√3×(2/√3)/((2x/√3-1/√3))²+1)-1/3×1/(x+1)
所以原积分
=∫[1/6×(2x-1)/(x²-x+1)+1/√3×(2/√3)/((2x/√3-1/√3))²+1)-1/3×1/(x+1)] dx
=(ln|x²-x+1|)/6 - (ln|x+1|)/3 + arctan((2x-1)/√3)/√3 + C