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)`

Since`u = 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))`