What is #f(x) = int x/sqrt(x^2-1) dx# if #f(3) = 0 #?

1 Answer
Eddie
Jul 2, 2016

#sqrt(x^2 - 1) - sqrt 8#

Explanation:

there's a very obvious pattern here

#f(x) = int dx qquad color(red)(x)/sqrt(color{red}{x^2}-1) #

we know that

#d/dx (y^(1/2)) = 1/2* y^(-1/2) * y'#

so we can trial estimate that if

#F(x) = alpha sqrt(x^2 - 1) #

then

#(dF)/dx = 1/2 alpha 1/sqrt(x^2 - 1) * 2x#

# = alpha \ x/sqrt(x^2 - 1) #

so #alpha = 1#

and

#f(x) = int dx qquad color(red)(x)/sqrt(color{red}{x^2}-1) #

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

you can of course play around with substitutions but seeing the pattern means that you can pretty much do the integration in your head. what i am really trying so say is that the integration is very trivial if you see the pattern, otherwise you need to get stuck into a lot of guessing and associated theory

it wold be interesting to see that array of subs that lead to an easy solution. i'd go for the simple #u^2 = x^2 -1# as an opening gambit

so, next, the initial value: #f(3) = 0#

#0 = sqrt 8 + C#

#implies f(x) = sqrt(x^2 - 1) - sqrt 8#

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