Calc1.17

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{\frac  {x!}{(x+n)!}}

={\frac  {1}{(x+1)(x+2)(x+3)...(x+n)}}

={\frac  {A_{1}}{x+1}}+{\frac  {A_{2}}{x+2}}+{\frac  {A_{3}}{x+3}}+...+{\frac  {A_{n}}{x+n}}

Multiplying through to create a common denominator then equating the numerators:

1={A_{1}}(x+2)(x+3)...(x+n)+{A_{2}}(x+1)(x+3)+...+(x+n)+{A_{3}}(x+1)(x+2)...(x+n)+...+{A_{n}}(x+1)(x+2)+...+(x+n-1)\,

Pick values of x\, to cancel out all but one term:

{\mbox{ Let }}x=-1,{\mbox{ then }}1={A_{1}}(1)(2)(3)...(n-1)\Rightarrow A_{1}={\frac  {1}{(n-1)!}}\,
{\mbox{ Let }}x=-2,{\mbox{ then }}1={A_{2}}(-1)(1)(2)...(n-2)\Rightarrow A_{2}={\frac  {1}{(-1)(n-2)!}}\,
{\mbox{ Let }}x=-3,{\mbox{ then }}1={A_{3}}(-2)(-1)(1)...(n-3)\Rightarrow A_{3}={\frac  {1}{(-2)(-1)(n-3)!}}\,

\vdots

{\mbox{ Let }}x=-n,{\mbox{ then }}1={A_{n}}(1-n)(2-n)(3-n)...(-1)\Rightarrow A_{n}={\frac  {\pm 1}{(n-1)!}}\,

Thus the general term for A_{r} can be deduced:

A_{r}={\frac  {(-1)^{{r-1}}}{(n-r)!(r-1)!}}\,

Hence \int {\frac  {x!}{(x+n)!}}\,dx=\int \sum _{{r=1}}^{n}{\frac  {A_{r}}{x+r}}\,dx=\sum _{{r=1}}^{n}A_{r}\ln |x+r|\,

=\sum _{{r=1}}^{n}{\frac  {(-1)^{{r-1}}\ln |x+r|}{(n-r)!(r-1)!}}