# Homework Help: Hard arc-length integral

1. Feb 24, 2013

### rhothi

Find the length of the curve
y = x^2 − ln(x), 1 ≤ x ≤ 2.

I have have to find the arc-length of the given equation. I have checked my work several times and i ultimately come up with the integral of ((4x^4 -3x^2 +1)/(x^2))^(1/2), which I just cannot seem to solve. Any help would be appreciated.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 24, 2013

### SammyS

Staff Emeritus
Hello rhothi. Welcome to PF !

3. Feb 24, 2013

### rhothi

I used the formula for calculating arc length. The formula is ((1 + (f'(x))^(2))^(1/2) from a to b. As a result, I got f'(x) to be 2x - 1/x. When I squared the function i received ((4x^4 -4x^2 +1). From there I plugged f'(x)^2 into the equation to get the integral that I mentioned.

4. Feb 24, 2013

### Karnage1993

$\displaystyle \left(2x - \frac{1}{x}\right)^2$ is not $4x^4 -4x^2 +1$.

Note that $(a+b)^2 = a^2 + 2ab + b^2$.

Try it again.

5. Feb 24, 2013

### rhothi

I meant that it is (4x^4−4x^2+1)/(x^2). After adding 1 underneath the radical I received ((4x^4 -3x^2 +1)/(x^2))^((1/2)).

6. Feb 24, 2013

### SammyS

Staff Emeritus
It looks like you have a typo.

Are you saying that $\displaystyle \ \ 1+\left(2x-\frac{1}{x}\right)^2=\frac{4x^4 -3x^2 +1}{x^2}\ ?$

That's correct.

The result for the integral, $\displaystyle \ \ \int\sqrt{\frac{4x^4 -3x^2 +1}{x^2}}\,dx\ \$ looks very messy.

You can solve for x as a function of y for x > 0 . That result doesn't look much more promising.

Off hand, I can't think of any parametrization which would give a nicer result.

7. Feb 24, 2013

### rhothi

Yes that is what I meant. How would I solve the equation in terms of y? I have no idea how to compute the integral in terms of x.

8. Feb 24, 2013

### SammyS

Staff Emeritus
Here is a parametrization which I thought looked promising, but not surprisingly, it gives a similarly messy result.

Let x = et, then y = (et) - ln(et) = e2t - t .

9. Feb 24, 2013

### Dick

If you put the integral into Wolfram Alpha, it does give you an indefinite integral in terms of elementary functions. I didn't think it would. That probably means you could do it by completing a square, thinking of a clever substitution, maybe doing an integration by parts or two and voila. But the complexity of the answer makes it look like a really tough road for mere mortals. I'd guess if this was an exercise in computing arclength, you want to make it a lot easier than that. I'm going to guess there is probably a typo in the problem statement.