請問 ∫1/[x(x + 1)(x + 2)· · ·(x + m)]dx 點計?

Let P(m)=∫1/[x(x + 1)(x + 2)· · ·(x + m)]dx.
m=1 ∫1/[x(x + 1)]dx= ∫1/xdx- ∫1/x+1)dx ==>P(1)=ln|x(x+1)^(-1)| +C;
m=2 ∫1/[x(x + 1)(x+2)]dx=(1/2){ ∫1/[x(x+1)]dx- ∫1/[(x+1)(x+2)}dx}, + P(1) above
==>P(2)=ln|x(x+1)^(-2)(x+2)|+C;
m=3 ∫1/[x(x + 1)(x+2)(x+3)]dx=(1/3){ ∫1/[x(x+1)(x+2)]dx- ∫1/[(x+1)(x+2)(x+3)}dx},
+ P(2) formula ==>P(3)=1/(3!)ln|x(x+1)^(-3)(x+2)^(3)(x+3)^(-1)| +C;
keep moving to the next m at one time, like evaluating the sum 1^k+2^k+....n^k , the general formula then arrives:
P(m)= [1/(m!)] ln|{[x^(C(m,0))][(x+1)^(-C(m,1)][(x+2)^(C(m,2))]...[(x+k)^[(-1)^kC(m.k)]...[(x+m)^[(-1)^mC(m,m)]}| +C, C: arbitrary constant.