Question

1/x的8次方×(1-x)的积分是多少呢

1/x的8次方×(1-x)的积分是多少呢

Answer 1

∫ dx/[x^8.(1-x)]
let
1/[x^8.(1-x)]
≡A1/x +A2/x^2+A3/x^3+A4/x^3+A5/x^3+A6/x^3+A7/x^3+A8/x^3 +B/(1-x)
=>
1≡A1.x^7.(1-x)+A2.x^6.(1-x)+A3.x^5.(1-x)+A4.x^4.(1-x)+A5.x^3.(1-x)
+A6.x^2.(1-x) +A7.x(1-x) +A8.(1-x) + Bx^8
x=1 => B=1
coef. of x^8 : -A1+B= 0 =>A1=1
coef. of x^7 : A1-A2= 0 =>A2=1
coef. of x^6 : A2-A3= 0 =>A3=1
...
A1=A2=...=A8=B=1
1/[x^8.(1-x)]
≡1/x +1/x^2+1/x^3+1/x^3+1/x^3+1/x^3+1/x^3+1/x^3 +1/(1-x)
∫ dx/[x^8.(1-x)]
=∫ [1/x +1/x^2+1/x^3+1/x^3+1/x^3+1/x^3+1/x^3+1/x^3 +1/(1-x)] dx
=ln|x| -1/x -(1/2)(1/x^2) -(1/3)(1/x^3)-(1/4)(1/x^4)-(1/5)(1/x^5)-(1/6)(1/x^6)
-(1/7)(1/x^7) - ln|1-x| + C