\(% 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 have by definition \[ \cosh a = \frac {e^a + e^{-a}}{2}. \]
Notice that \begin {align*} \frac {1}{x^2 + 2x \cosh a + 1} & = \frac {1}{x^2 + (e^a + e^{-a}) x + (e^a \cdot e^{-a})} \\ & = \frac {1}{(x + e^a) (x + e^{-a})} \\ & = \left (\frac {1}{x + e^{-a}} - \frac {1}{x + e^{a}}\right ) \cdot \frac {1}{e^{a} - e^{-a}}, \end {align*}
and hence \[ \int \frac {\Diff x}{x^2 + 2x \cosh a + 1} = \frac {\ln \abs *{x + e^{-a}} - \ln \abs *{x + e^a}}{e^a - e^{-a}} = \frac {1}{e^a - e^{-a}} \ln \abs *{\frac {x + e^{-a}}{x + e^a}}. \]
Therefore, \begin {align*} \int _{0}^{1} \frac {\Diff x}{x^2 + 2x \cosh a + 1} & = \frac {1}{e^a - e^{-a}} \left [\ln \abs *{\frac {1 + e^{-a}}{1 + e^a}} - \ln \abs *{\frac {e^{-a}}{e^a}}\right ] \\ & = \frac {1}{e^a - e^{-a}} \left [\ln \abs *{\frac {1 + e^{-a}}{e^a \left (1 + e^{-a}\right )}} + 2a\right ] \\ & = \frac {1}{e^a - e^{-a}} \left [-a + 2a\right ] \\ & = \frac {a}{e^a - e^{-a}} \\ & = \frac {a}{2 \sinh a}. \end {align*}
For the first integral, we have by definition \[ \sinh a = \frac {e^a - e^{-a}}{2}. \]
Notice that \begin {align*} \frac {1}{x^2 + 2x \sinh a - 1} & = \frac {1}{x^2 + (e^a - e^{-a})x - (e^a \cdot e^{-a})} \\ & = \frac {1}{(x + e^a) (x - e^{-a})} \\ & = \left (\frac {1}{x - e^{-a}} - \frac {1}{x + e^a}\right ) \cdot \frac {1}{e^a + e^{-a}}, \end {align*}
and hence \[ \int \frac {\Diff x}{x^2 + 2x \sinh a - 1} = \frac {\ln \abs *{x - e^{-a}} - \ln \abs *{x + e^a}}{e^a + e^{-a}} = \frac {1}{e^a + e^{-a}} \ln \abs *{\frac {x - e^{-a}}{x + e^a}}. \]
Therefore, \begin {align*} \int _{1}^{\infty } \frac {\Diff x}{x^2 + 2x \sinh a - 1} & = \frac {1}{e^a + e^{-a}} \cdot \left [\ln \abs *{\frac {x - e^{-a}}{x + e^a}}\right ]_{1}^{\infty } \\ & = \frac {1}{2 \cosh a} \cdot \left [\ln 1 - \ln \abs *{\frac {1 - e^{-a}}{1 + e^a}}\right ] \\ & = \frac {1}{2 \cosh a} \cdot \ln \frac {1 + e^a}{1 - e^{-a}} \\ & = \frac {1}{2 \cosh a} \cdot \left (a + \ln \frac {1 + e^{-a}}{1 - e^{-a}}\right ) \\ & = \frac {1}{2 \cosh a} \cdot \left (a + \ln \frac {e^{\frac {a}{2}} + e^{-\frac {a}{2}}}{e^{\frac {a}{2}} - e^{-\frac {a}{2}}}\right ) \\ & = \frac {1}{2 \cosh a} \cdot \left (a + \ln \coth \frac {a}{2}\right ). \end {align*}
For the second integral, notice that \begin {align*} \frac {1}{x^4 + 2x^2 \cosh a + 1} & = \frac {1}{e^a - e^{-a}} \left (\frac {1}{x^2 + e^{-a}} - \frac {1}{x^2 + e^{a}}\right ), \end {align*}
and hence \begin {align*} \int _{0}^{\infty } \frac {\Diff x}{x^4 + 2x^2 \cosh a + 1} & = \frac {1}{2 \sinh a} \int _{0}^{\infty } \left (\frac {1}{x^2 + e^{-a}} - \frac {1}{x^2 + e^{a}}\right ) \Diff x \\ & = \frac {1}{2 \sinh a} \left [e^{\frac {a}{2}} \arctan \left (e^{\frac {a}{2}} x\right ) - e^{-\frac {a}{2}} \arctan \left (e^{-\frac {a}{2}} x\right ) \right ]_{0}^{\infty } \\ & = \frac {1}{2 \sinh a} \left [\left (e^{\frac {a}{2}} \frac {\pi }{2} - e^{-\frac {a}{2}} \frac {\pi }{2}\right ) - \left (e^{\frac {a}{2}} 0 - e^{-\frac {a}{2}} 0\right )\right ] \\ & = \frac {1}{2 \sinh a} \cdot (e^{\frac {a}{2}} - e^{-\frac {a}{2}}) \cdot \frac {\pi }{2} \\ & = \frac {\pi \sinh \frac {a}{2}}{2 \sinh a} \\ & = \frac {\pi \sinh \frac {a}{2}}{4 \sinh \frac {a}{2} \cosh \frac {a}{2}} \\ & = \frac {\pi }{4 \cosh \frac {a}{2}}. \end {align*}