\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We notice that \begin {align*} \RHS & = (x - y) \sum _{r = 1}^{n} x^{n - r} y^{r - 1} \\ & = x \sum _{r = 1}^{n} x^{n - r} y^{r - 1} - y \sum _{r = 1}^{n} x^{n - r} y^{r - 1} \\ & = \sum _{r = 1}^{n} x^{n - r + 1} y^{r - 1} - \sum _{r = 1}^{n} x^{n - r} y^{r} \\ & = \sum _{r = 0}^{n - 1} x^{n - r} y^r - \sum _{r = 1}^{n} x^{n - r} y^{r} \\ & = x^n y^0 + \sum _{r = 1}^{n - 1} x^{n - r} y^r - \sum _{r = 1}^{n - 1} x^{n - r} y^{r} - x^0 y^n \\ & = x^n - y^n. \end {align*}
Notice that \begin {align*} f(x) & = x^n \cdot \left (F(x) - \frac {A}{x - k}\right ) \\ & = x^n \cdot \left (\frac {1}{x^n (x - k)} - \frac {A}{x - k}\right ) \\ & = \frac {1}{x - k} - \frac {A x^n}{x - k} \\ & = \frac {1 - A x^n}{x - k}. \end {align*}
Since \(f\) is a polynomial, the numerator must be divisible by the denominator, and hence when \(x = k\), the numerator must be \(0\), which means \[ 1 - A k^n = 0, \] and hence \[ A = \frac {1}{k^n}. \]
Hence, \[ f(x) = \frac {1 - A x^n}{x - k} = \frac {1}{x - k} \left (1 - \left (\frac {x}{k}\right )^n\right ), \] as desired.
Using the identity, we have \begin {align*} f(x) & = \frac {1 - A x^n}{x - k} \\ & = \frac {1^n - \left (\frac {x}{k}\right )^n}{x - k} \\ & = \frac {1}{k^n} \cdot \frac {k^n - x^n}{x - k} \\ & = \frac {1}{k^n} \cdot \frac {-(x - k) \sum _{r = 1}^{n} k^{n - r} x^{r - 1}}{x - k} \\ & = - \sum _{r = 1}^{n} k^{-r} x^{r - 1}, \end {align*}
and hence \begin {align*} F(x) & = \frac {A}{x - k} + \frac {f(x)}{x^n} \\ & = \frac {1}{k^n (x - k)} - \frac {\sum _{r = 1}^{n} k^{-r} x^{r - 1}}{x^n} \\ & = \frac {1}{k^n (x - k)} - \sum _{r = 1}^{n} \frac {1}{k^r x^{n - r + 1}} \\ & = \frac {1}{k^n (x - k)} - \sum _{r = 1}^{n} \frac {1}{k^{n - r + 1} x^r} \\ & = \frac {1}{k^n (x - k)} - \frac {1}{k} \sum _{r = 1}^{n} \frac {1}{k^{n - r} x^r}, \end {align*}
as desired.
Notice that on one hand, \[ \DiffOp {x} x^n F(x) = \DiffOp {x} \frac {1}{x - k} = - \frac {1}{(x - k)^2}, \] and on the other hand, using the expression above, we have \begin {align*} \DiffOp {x} x^n F(x) & = \DiffOp {x} \left [\frac {x^n}{k^n (x - k)} - \frac {1}{k} \sum _{r = 1}^{n} \frac {x^{n - r}}{k^{n - r}}\right ] \\ & = \frac {n x^{n - 1} k^n (x - k) - x^n k^n}{k^{2n} (x - k)^2} - \frac {1}{k} \sum _{r = 1}^{n} \frac {(n - r) x^{n - r - 1}}{k^{n - r}} \\ & = \frac {n x^{n - 1} (x - k) - x^n}{k^n (x - k)^2} - \sum _{r = 1}^{n} \frac {n - r}{k^{n - r + 1} x^{-n + r + 1}} \\ & = \frac {n x^{n - 1}}{k^n (x - k)} - \frac {x^n}{k^n (x - k)^2} - \sum _{r = 1}^{n} \frac {n - r}{k^{n - r + 1} x^{-n + r + 1}}. \end {align*}
Hence, \begin {align*} -\frac {1}{(x - k)^2} & = \frac {n x^{n - 1}}{k^n (x - k)} - \frac {x^n}{k^n (x - k)^2} - \sum _{r = 1}^{n} \frac {n - r}{k^{n - r + 1} x^{-n + r + 1}} \\ \frac {1}{(x - k)^2} & = \frac {x^n}{k^n (x - k)^2} - \frac {n x^{n - 1}}{k^n (x - k)} + \sum _{r = 1}^{n} \frac {n - r}{k^{n - r + 1} x^{-n + r + 1}} \\ \frac {1}{x^n (x - k)^2} & = \frac {1}{k^n (x - k)^2} - \frac {n}{k^n x (x - k)} + \sum _{r = 1}^{n} \frac {n - r}{k^{n - r + 1} x^{r + 1}}, \end {align*}
precisely as desired.
Let \(n = 3\) and \(k = 1\), and hence we have \begin {align*} \frac {1}{x^3 (x - 1)^2} & = \frac {1}{(x - 1)^2} - \frac {3}{x (x - 1)} + \sum _{r = 1}^{3} \frac {3 - r}{x^{r + 1}} \\ & = \frac {1}{(x - 1)^2} - \frac {3}{x - 1} + \frac {3}{x} + \frac {2}{x^2} + \frac {1}{x^3}. \end {align*}
Hence, we have \begin {align*} \int _{2}^{N} \frac {\Diff x}{x^3 (x - 1)^2} & = \int _{2}^{N} \left [\frac {1}{(x - 1)^2} - \frac {3}{x - 1} + \frac {3}{x} + \frac {2}{x^2} + \frac {1}{x^3}\right ] \Diff x \\ & = \left [- \frac {1}{x - 1} - 3 \ln \abs *{x - 1} + 3 \ln \abs *{x} - \frac {2}{x} - \frac {1}{2x^2}\right ]_{2}^{N} \\ & = \left (3 \ln \frac {N}{N - 1} - \frac {1}{N - 1} - \frac {2}{N} - \frac {1}{2 N^2}\right ) - \left (3 \ln \frac {2}{1} - \frac {1}{1} - \frac {2}{2} - \frac {1}{2 \cdot 4}\right ) \\ & = \frac {17}{8} - 3 \ln 2 + \left (3 \ln \frac {N}{N - 1} - \frac {1}{N - 1} - \frac {2}{N} - \frac {1}{2 N^2}\right ). \end {align*}
We take the limit as \(N \to \infty \). Since \(\frac {N}{N - 1} = 1 + \frac {1}{N - 1}\), and as \(N \to \infty \), \(\frac {1}{N - 1} \to 0\), this means that \(\ln \frac {N}{N - 1} \to 0\). All the fractions with \(N\) on the denominator also approaches 0. Hence, the limit of this integral as \(N \to \infty \) is \[ \int _{2}^{\infty } \frac {\Diff x}{x^3 (x - 1)^2} = \frac {17}{8} - 3 \ln 2. \]