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?

\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.
