Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 100

To prove
sinh^-1 t =ln(t+sqrt(t^2+1))
let
sinh^-1 t = x
t=sinh(x)= (e^x -e^(-x))/2 =(e^x - (1/(e^x)))/2= (e^(2x) -1)/(2(e^x))
let t= (e^(2x) -1)/(2(e^x))
=> 2e^x t = e^(2x) -1
=> let e^x = u so,
2ut=u^2 -1
=> u^2 -2ut -1 =0 is of the quadratic form ax^2 +bx+c = 0 so finding the roots using the quadratic formula
(-b+-sqrt(b^2 -4ac))/(2a)
here in the equation u^2 -2ut -1 =0
a=1 , b=-2, c=-1
u=(-(-2t)+-sqrt(4t^2-4(1)(-1)))/2
u=(2t+-sqrt(4t^2+4))/2
=(2t+-2sqrt(t^2+1))/2
=t+-sqrt(t^2+1)
Sinceu = e^x > 0 then t+sqrt(t^2+1)>0
So e^x=t+sqrt(t^2+1)
x=ln(t+sqrt(t^2+1))
Since
sinh^-1 t = x
it follows that
sinh^-1 t = ln(t+sqrt(t^2+1))