Expert: Paul Klarreich - 3/16/2009

Question
Differentiate the Function :
h(x)=ln(x+SQUARE ROOT(x^2-1)


Evaluate the Integral from 1to2 using natural log:
(4+u^2/u^3)

Answer
Questioner:   Rachel Guyer
Category:  Calculus
Private:  No
 
Subject:  Calculus Question
Question:  Differentiate the Function :
h(x)=ln(x+SQUARE ROOT(x^2-1)    <<< needs a paren.


Evaluate the Integral from 1to2 using natural log:
(4+u^2/u^3)
--------------------------------------------  

h(x) = ln(x + sqrt((x^2-1))

Chain rule:  
let  u = x + sqrt(x^2 - 1)

du            x
-- = 1 + -------------   << you work this out.
dx       sqrt(x^2 - 1)


du    sqrt(x^2 - 1) + x
-- =  -----------------   << combine over a LCD
dx      sqrt(x^2 - 1)

du    x + sqrt(x^2 - 1)   << just a quick switch.
-- =  -----------------
dx      sqrt(x^2 - 1)


dy    1  du
-- = --- --
du    u  dx

         1               x + sqrt(x^2 - 1)
= --------------------- --------------------
 x + sqrt(x^2 - 1)        sqrt(x^2 - 1)


         1
= -----------------
  sqrt(x^2 - 1)

Check this at:http://integrals.wolfram.com/index.jsp

.................................................
Evaluate the Integral from 1to2 using natural log: << PLEASE USE THE SPACE BAR
(4+u^2/u^3)


{2  4 + u^2
|  -------- du   << you need the du here.
}1    u^3

{2  
|  (4u^-3 + u^-1) du =
}1

4u^-2
----- + ln u =
-2

- 2
----- + ln u  from  1 to 2 =
u^2

(-2/4 + ln 2) - (-2 + ln 1) =

-1/2 + ln 2 + 2  = etc.