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

Jul 2, 2016

$\sqrt{{x}^{2} - 1} - \sqrt{8}$

#### Explanation:

there's a very obvious pattern here

$f \left(x\right) = \int \mathrm{dx} q \quad \frac{\textcolor{red}{x}}{\sqrt{\textcolor{red}{{x}^{2}} - 1}}$

we know that

$\frac{d}{\mathrm{dx}} \left({y}^{\frac{1}{2}}\right) = \frac{1}{2} \cdot {y}^{- \frac{1}{2}} \cdot y '$

so we can trial estimate that if

$F \left(x\right) = \alpha \sqrt{{x}^{2} - 1}$

then

$\frac{\mathrm{dF}}{\mathrm{dx}} = \frac{1}{2} \alpha \frac{1}{\sqrt{{x}^{2} - 1}} \cdot 2 x$

$= \alpha \setminus \frac{x}{\sqrt{{x}^{2} - 1}}$

so $\alpha = 1$

and

$f \left(x\right) = \int \mathrm{dx} q \quad \frac{\textcolor{red}{x}}{\sqrt{\textcolor{red}{{x}^{2}} - 1}}$

$= \sqrt{{x}^{2} - 1} + C$

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: $f \left(3\right) = 0$

$0 = \sqrt{8} + C$

$\implies f \left(x\right) = \sqrt{{x}^{2} - 1} - \sqrt{8}$