∫x(x^8-1)dx 详细过程!谢谢!-dx8.1
编辑： admin 2017-01-03
-
4
答：
∫x/(x^8-1)dx
=∫1/2*(x/(x^4-1)-x/(x^4+1))dx
=1/2∫1/2*(x/(x^2-1)-x/(x^2+1))dx-1/2∫x/(x^4+1)dx
=1/4∫x/(x^2-1)dx-1/4∫x/(x^2+1)dx-1/2∫x/(x^4+1)dx
=1/8∫1/(x^2-1)d(x^2-1)-1/8∫1/(x^2+1)d(x^2+1)-1/4∫1/(x^4+1)d(x^2)
=1/8[ln|x^2-1|-ln(x^2+1)]-1/4arctanx^2+C
=1/8ln|(x^2-1)/(x^2+1)|-1/4arctanx^2+C
类似问题
类似问题1：∫1/[x-√(1-x^2)]dx[数学科目]
令x=sint,dx=cost
∫1/[x-√(1-x^2)]dx
=∫cost/(sint-cost)dt
令cost=a[sint-cost]+b[sint-cost]'=a[sint-cost]+b[cost+sint]=(a+b)sint+(b-a)cost
则a+b=0,b-a=1,解得a=-1/2,b=1/2
那么∫cost/(sint-cost)dt
=(1/2)∫[-(sint-cost)+(sint-cost)']/(sint-cost)dt
=-(1/2)∫dt+(1/2)∫1/(sint-cost)d(sint-cost)
=1/2[-t+ln|sint-cost|+C
三角代换t=arcsinx,sint=x,cost=√(1-x²)
所以
∫1/[x-√(1-x^2)]dx
=1/2[-arcsinx+ln|x-√(1-x²)|]+C
类似问题2：∫1/(x^2-4x+8)dx,求不定积分,[数学科目]
类似问题3：1/(x+x^2)dx[数学科目]
∫1/(x+x^2)dx
=∫1/xdx -∫1/(x+1)dx
=ln|x|-ln|x+1| +C
类似问题4：求∫1/1+√x dx[数学科目]
令t=√x,那么x=t²,原式dx=d(t²)=2tdt
原式=∫1/(1+t)*2tdt
=∫2t/(1+t)dt
=∫(2t+2-2)/(1+t)dt
=∫[2-2/(1+t)]dt
=2t-2ln(1+t)+C
=2√x-2ln(1+√x)+C
望采纳
类似问题5：求积分 ∫1到8 [(x^1/3+1)(2-x^2/3)]/x^1/3 dx[数学科目]
=∫[x^(1/3)-x+2-x^(2/3)]/x^(1/3)]dx
=∫[1-x^(2/3)+2x^(-1/3)-x^(1/3)]]dx
幂函数积分
=x-x^(5/3)/(5/3)+2x^(2/3)/(2/3)-x^(4/3)/(4/3)(1,8)
代入算一下吧