# Compute the exact length of the curve $y = \frac{x^2}{4}-\frac{1}{2}\ln(x)$ from $x=1, x=2$

So here is my problem: $y = \frac{x^2}{4}-\frac{1}{2}\ln(x)$

Taking the derivative:

$$\frac{\textrm{d}y}{\textrm{d}x}=\frac{x}{2}-\frac{1}{2x}$$

And that simplifies further to:

$$\frac{\textrm{d}y}{\textrm{d}x}=\frac{x^2-1}{2x}$$

Since the formula for the curve is $$\int \sqrt{1+\left(\frac{\textrm{d}y}{\textrm{d}x}\right)^2}\textrm{d}x$$ I know I have to square my derivative:

$$1+\left(\frac{\textrm{d}y}{\textrm{d}x}\right)^2=1+\left(\frac{x^2-1}{2x}\right)^2$$

After expanding and adding $1$, I got the following:

$$\frac{5x^2-2x+1}{4x^2}$$

And I don't know what to do with it from here since the numerator doesn't seem to factor into a perfect square. Any suggestions?

• You didn't square the numerator correctly. – cygorx Nov 6 '13 at 4:01
• @cygorx I don't see where I messed up in squaring the numerator? – hax0r_n_code Nov 6 '13 at 4:06
• @zibadawatimmy so I can square the numerator and denominator separately since they share common denominator? – hax0r_n_code Nov 6 '13 at 4:08
• Well, yes. $(ab)^2 = a^2 b^2$ (until you deal with things like matrices, but in this context $ab=ba$ holds). – zibadawa timmy Nov 6 '13 at 4:09
• Might I add that this integral works out quite nicely. – cygorx Nov 6 '13 at 4:12

$$l = \left[\frac{4}{4}+\frac{\ln\left(2\right)}{2}\right] - \left(\frac{1}{4} + 0\right) = \frac{3}{4}+\frac{\ln\left(2\right)}{2}$$
• @ThanosDarkadakis I didn't change anything you wrote. I just add some $\tt LaTeX$. Read the edit and you'll find some advice. Also, you can write your macros as enclosed in . That's is very useful. That was a fine answer. – Felix Marin Nov 6 '13 at 5:56