\(% 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}}\)
By Newton’s binomial theorem, we have \begin {align*} (8 + x^3)^{-1} & = \frac {1}{8} \left (1 + \left (\frac {x}{2}\right )^3\right )^{-1} \\ & = \frac {1}{8} \sum _{k = 0}^{\infty } (-1)^k \left (\frac {x}{2}\right )^{3k}, \end {align*}
and this is valid for \[ \abs *{\frac {x}{2}} < 1, \abs *{x} < 2, \] as desired.
Hence, \begin {align*} \int _{0}^{1} \frac {x^m}{8 + x^3} \Diff x & = \int _{0}^{1} \frac {1}{8} \sum _{k = 0}^{\infty } (-1)^k \left (\frac {x}{2}\right )^{3k} x^m \Diff x \\ & = \frac {1}{8} \sum _{k = 0}^{\infty } \frac {(-1)^k}{2^{3k}} \int _{0}^{1} x^{3k + m} \Diff x \\ & = \sum _{k = 0}^{\infty } \frac {(-1)^k}{2^{3(k + 1)}} \left [\frac {x^{3k + m + 1}}{3k + m + 1}\right ]_{0}^{1} \\ & = \sum _{k = 0}^{\infty } \left (\frac {(-1)^k}{2^{3(k + 1)}} \cdot \frac {1}{3k + m + 1}\right ), \end {align*}
as desired.
Let \(m = 2\), and we have \[ \int _{0}^{1} \frac {x^2}{8 + x^3} \Diff x = \sum _{k = 0}^{\infty } \left (\frac {(-1)^k}{2^{3(k + 1)}} \cdot \frac {1}{3k + 3}\right ). \]
Let \(m = 1\), and we have \[ \int _{0}^{1} \frac {x}{8 + x^3} \Diff x = \sum _{k = 0}^{\infty } \left (\frac {(-1)^k}{2^{3(k + 1)}} \cdot \frac {1}{3k + 2}\right ). \]
Let \(m = 0\), and we have \[ \int _{0}^{1} \frac {x}{8 + x^3} \Diff x = \sum _{k = 0}^{\infty } \left (\frac {(-1)^k}{2^{3(k + 1)}} \cdot \frac {1}{3k + 1}\right ). \]
Hence, \begin {align*} \sum _{k = 0}^{\infty } \frac {(-1)^k}{2^{3(k + 1)}} \left (\frac {1}{3k + 3} - \frac {2}{3k + 2} + \frac {4}{3k + 1}\right ) & = \int _{0}^{1} \frac {x^2}{8 + x^3} \Diff x - 2 \int _{0}^{1} \frac {x}{8 + x^3} \Diff x + 4 \int _{0}^{1} \frac {\Diff x}{8 + x^3} \\ & = \int _{0}^{1} \frac {x^2 - 2x + 4}{8 + x^3} \Diff x \\ & = \int _{0}^{1} \frac {x^2 - 2x + 4}{(x + 2)(x^2 - 2x + 4)} \Diff x \\ & = \int _{0}^{1} \frac {\Diff x}{x + 2} \\ & = \left [\ln \abs *{x + 2}\right ]_{0}^{1} \\ & = \ln 3 - \ln 2. \end {align*}
Using partial fractions, let \(A'\) and \(B'\) be real constants such that \begin {align*} \frac {72(2k + 1)}{(3k + 1)(3k + 2)} & = \frac {A'}{3k + 1} + \frac {B'}{3k + 2} \\ & = \frac {3(A' + B')k + (2A' + B')}{(3k + 1) (3k + 2)}. \end {align*}
Hence, we have \[ \left \{\begin {aligned} 3(A' + B') & = 72 \cdot 2 = 144, \\ 2A' + B' & = 72. \end {aligned}\right . \]
Therefore, \((A', B') = (24, 24)\).
Let \[ A = \int _{0}^{1} \frac {\Diff x}{8 + x^3}, B = \int _{0}^{1} \frac {x \Diff x}{8 + x^3}, C = \int _{0}^{1} \frac {x^2 \Diff x}{8 + x^3}, \] and what is desired is \(24(A + B)\).
From the previous part, we can see that \(4A - 2B + C = \ln 3 - \ln 2\).
Also, \begin {align*} 2A + B & = \int _{0}^{1} \frac {(2 + x) \Diff x}{8 + x^3} \\ & = \int _{0}^{1} \frac {\Diff x}{x^2 - 2x + 4} \\ & = \int _{0}^{1} \frac {\Diff x}{(x - 1)^2 + 3} \\ & = \frac {1}{\sqrt {3}} \left [\arctan \left (\frac {x - 1}{\sqrt {3}}\right )\right ]_{0}^{1} \\ & = \frac {1}{\sqrt {3}} \cdot \left [\arctan 0 - \arctan \left (- \frac {1}{\sqrt {3}}\right )\right ] \\ & = \frac {1}{\sqrt {3}} \cdot \frac {\pi }{6} \\ & = \frac {\pi }{6\sqrt {3}}. \end {align*}
We also have \begin {align*} C & = \int _{0}^{1} \frac {x^2 \Diff x}{8 + x^3} \\ & = \frac {1}{3} \left [\ln (8 + x^3)\right ]_{0}^{1} \\ & = \frac {1}{3} \left [\ln 9 - \ln 8\right ] \\ & = \frac {2}{3} \ln 3 - \ln 2. \end {align*}
Hence, we have \[ 4A - 2B = \ln 3 - \ln 2 - \frac {2}{3} \ln 3 + \ln 2 = \frac {1}{3} \ln 3, \] and hence \(2A - B = \frac {1}{6} \ln 3\).
Therefore, \[ 4A = \frac {1}{6} \ln 3 + \frac {\pi }{6\sqrt {3}}, \] and hence \[ A = \frac {\ln 3}{24} + \frac {\pi }{24\sqrt {3}}. \]
Subtracting two of this from \(2A + B\) gives \[ B = \frac {\pi }{6 \sqrt {3}} - \frac {\ln 3}{12} - \frac {\pi }{12 \sqrt {3}} = \frac {\pi }{12\sqrt {3}} - \frac {\ln 3}{12}, \] and hence what is desired is \begin {align*} 24(A + B) & = 24 \left (\frac {\pi }{24 \sqrt {3}} + \frac {\pi }{12 \sqrt {3}} + \frac {\ln 3}{24} - \frac {\ln 3}{12}\right ) \\ & = 24 \left (\frac {\pi }{8 \sqrt {3}} - \frac {\ln 3}{24}\right ) \\ & = \pi \sqrt {3} - \ln 3, \end {align*}
which gives \(a = 3, b = 3\).