Primitive of Inverse Hyperbolic Secant of x over a over x/Lemma

Lemma for Primitive of Inverse Hyperbolic Secant of x over a over x

$\map {\ln^2} {\dfrac x {2 a} } = \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x} + \map {\ln^2} 2$

Proof

$$\ln \left(\frac{a}{x}\right) \ln \left(\frac{4 a}{x}\right)-\ln ^2\left(\frac{x}{2 a}\right)= \ln \left(\frac{a}{x}\right) \left[\ln \left(\frac{a}{x}\right)+2 \log (2)\right]-\left[\ln \left(\frac{a}{x}\right)+\ln (2)\right]^2 =-\ln ^2(2)$$

\(\ds \map {\ln^2} {\dfrac x {2 a} }\)

\(=\)

\(\ds \paren {\map \ln {\dfrac x {2 a} } }^2\)

\(\ds \)

\(=\)

\(\ds \paren {-\map \ln {\dfrac {2 a} x} }^2\)

Logarithm of Reciprocal

\(\ds \)

\(=\)

\(\ds \paren {\map \ln {\dfrac {2 a} x} }^2\)

\(\ds \)

\(=\)

\(\ds \paren {\map \ln {\dfrac a x} + \ln 2}^2\)

\(\ds \)

\(=\)

\(\ds \paren {\map \ln {\dfrac a x} }^2 + 2 \ln 2 \map \ln {\dfrac a x} + \paren {\ln 2}^2\)

Then:

\(\ds \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x}\)

\(=\)

\(\ds \map \ln {\dfrac a x} \paren {\map \ln {\dfrac a x} + \map \ln {2^2} }\)

Sum of Logarithms

\(\ds \)

\(=\)

\(\ds \map \ln {\dfrac a x} \paren {\map \ln {\dfrac a x} + 2 \ln 2}\)

\(\ds \)

\(=\)

\(\ds \paren {\map \ln {\dfrac a x} }^2 + 2 \ln 2 \map \ln {\dfrac a x}\)

Hence the result.

$\blacksquare$