\(% 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 show that \(T\) is equal to each of \(U, V, X\), and by transitivity, this shows that all four are equal.
To show \(T = U\), consider the substitution \(u = 2 \artanh t\), and hence \(t = \tanh \frac {u}{2}\).
When \(t = \frac {1}{2}\), \(u = 2 \artanh \frac {1}{2} = 2 \cdot \frac {1}{2} \cdot \ln \left (\frac {1 + \frac {1}{2}}{1 - \frac {1}{2}}\right ) = \ln 3\), and when \(t = \frac {1}{3}\), \(u = 2 \artanh \frac {1}{3} = 2 \cdot \frac {1}{2} \cdot \ln \left (\frac {1 + \frac {1}{3}}{1 - \frac {1}{3}}\right ) = \ln 2\).
We have \(\Diff u = \frac {2}{1 - t^2} \Diff t\), and hence \begin {align*} T & = \int _{\frac {1}{3}}^{\frac {1}{2}} \frac {\artanh t}{t} \Diff t \\ & = \int _{\ln 2}^{\ln 3} \frac {\frac {u}{2}}{\tanh \frac {u}{2}} \cdot \frac {1 - \tanh ^2 \frac {u}{2}}{2} \Diff u \\ & = \int _{\ln 2}^{\ln 3} \frac {u}{2} \cdot \frac {1 - \tanh ^2 \frac {u}{2}}{2 \tanh \frac {u}{2}} \Diff u \\ & = \int _{\ln 2}^{\ln 3} \frac {u}{2 \sinh u} \Diff u \\ & = U. \end {align*}
To show \(T = X\), consider the substitution \(x = - \frac {1}{2} \ln t\), and hence \(t = e^{-2x}\).
When \(t = \frac {1}{2}\), \(x = - \frac {1}{2} \ln \frac {1}{2} = \frac {1}{2} \ln 2\), and when \(t = \frac {1}{3}\), \(x = - \frac {1}{2} \ln \frac {1}{3} = \frac {1}{2} \ln 3\).
We have \(\Diff x = - \frac {\Diff t}{2t}\), and hence \begin {align*} T & = \int _{\frac {1}{3}}^{\frac {1}{2}} \frac {\artanh t}{t} \Diff t \\ & = \int _{\frac {1}{2} \ln 3}^{\frac {1}{2} \ln 2} \frac {\artanh e^{-2x}}{t} \cdot (-2t) \Diff x \\ & = \int _{\frac {1}{2} \ln 2}^{\frac {1}{2} \ln 3} 2 \artanh e^{-2x} \Diff x \\ & = \int _{\frac {1}{2} \ln 2}^{\frac {1}{2} \ln 3} \ln \left (\frac {1 + e^{-2x}}{1 - e^{-2x}}\right ) \Diff x \\ & = \int _{\frac {1}{2} \ln 2}^{\frac {1}{2} \ln 3} \ln \left (\frac {e^{x} + e^{-x}}{e^{x} - e^{-x}}\right ) \Diff x \\ & = \int _{\frac {1}{2} \ln 2}^{\frac {1}{2} \ln 3} \ln \coth x \Diff x \\ & = X. \end {align*}