Let #sech^(-1)x=t# then #x=secht#,
then #x=1/cosht=2/(e^t+e^(-t)#
or #e^t+e^(-t)=2/x#
or #e^t+1/e^t=2/x#
or #(e^t)^2-2/xe^t+1=0#
Solving for #e^t#, #e^t=(-(-2/x)+-sqrt(4/x^2-4))/2#
= #1/x+-1/xsqrt(1-x^2)#
When #x>0#, we have #e^t=1/x+1/xsqrt(1-x^2)=(1+sqrt(1-x^2))/x#
and taking log we have #t=sech^(-1)x=ln((1+sqrt(1-x^2))/x)#
and #sech^(-1)x+lnx=3/2#
or #ln((1+sqrt(1-x^2))/x)+lnx=ln(3/2)#
or #ln(1+sqrt(1-x^2))=ln(3/2)#
or #1+sqrt(1-x^2)=3/2#
or #sqrt(1-x^2)=1/2#
or #1-x^2=1/4#
or #x^2-3/4=0# i.e. #x=+-sqrt3/2#
but we have #x>0# hence #x=sqrt3/2#
