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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?

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    $\begingroup$ You didn't square the numerator correctly. $\endgroup$ – cygorx Nov 6 '13 at 4:01
  • $\begingroup$ @cygorx I don't see where I messed up in squaring the numerator? $\endgroup$ – hax0r_n_code Nov 6 '13 at 4:06
  • $\begingroup$ @zibadawatimmy so I can square the numerator and denominator separately since they share common denominator? $\endgroup$ – hax0r_n_code Nov 6 '13 at 4:08
  • $\begingroup$ Well, yes. $(ab)^2 = a^2 b^2$ (until you deal with things like matrices, but in this context $ab=ba$ holds). $\endgroup$ – zibadawa timmy Nov 6 '13 at 4:09
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    $\begingroup$ Might I add that this integral works out quite nicely. $\endgroup$ – cygorx Nov 6 '13 at 4:12
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\begin{align} &\int_{1}^{2}\sqrt{1 + \left(\frac{{\rm d}y}{{\rm d}x}\right)^2}\,{\rm d}x = \int_{1}^{2}\sqrt{1 + \left(\frac{x^{2} - 1}{2x}\right)^2}\,{\rm d}x = \int_{1}^{2}\sqrt{1 + \frac{x^{4} - 2x^{2} + 1}{4x^{2}}}\,{\rm d}x \\[3mm]&= \int_{1}^{2}\sqrt{\frac{x^{4} + 2x^{2} + 1}{4x^{2}}}\,{\rm d}x = \int_{1}^{2}\sqrt{\left(\frac{x^{2} + 1}{2x}\right)^{2}}\,{\rm d}x = \int_{1}^{2}{\frac{x^{2} + 1}{2x}}\,{\rm d}x = \int_{1}^{2}\left(\frac{x}{2} + \frac{1}{2x}\right)\,{\rm d}x \\[3mm]&= \left[\frac{x^2}{4}+\frac{\ln\left(x\right)}{2}\right]_{1}^{2} \end{align}

By applying your limits we get:

$$ l = \left[\frac{4}{4}+\frac{\ln\left(2\right)}{2}\right] - \left(\frac{1}{4} + 0\right) = \frac{3}{4}+\frac{\ln\left(2\right)}{2} $$

Sorry but it is my first post. took me so much time to learn to use the symbols.

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  • $\begingroup$ You'll catch on. Look at how other people format their answers by right clicking on the MathJax and choosing "Show Math As > TeX Commands" (+1) $\endgroup$ – robjohn Nov 6 '13 at 5:02
  • $\begingroup$ Oh thanks. all this time i was using edit.. $\endgroup$ – Thanos Darkadakis Nov 6 '13 at 5:04
  • $\begingroup$ @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. $\endgroup$ – Felix Marin Nov 6 '13 at 5:56

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