Arc length of $f(x)=x^2-\ln x$ over [1,e] This is how I learned to solve for arc length:
$$ \frac d{dx} f(x)=2x - \frac 1x$$
$$ \left(\frac d{dx}f(x)\right)^2 = \left(2x - \frac 1x\right)^2 = 4x^2-4+\frac1{x^2} $$
$$ 1+\left(\frac d{dx}f(x)\right)^2 = 4x^2-3+\frac1{x^2} $$
$$ \sqrt{1+\left(\frac d{dx}f(x)\right)^2} = \sqrt{4x^2-3+\frac1{x^2}}$$
So to solve for arc length, I have:
$$\int_1^e \sqrt{4x^2-3+\frac1{x^2}} ~dx $$
I already have the answer (from Wolfram alpha) but I cannot figure out how to solve this integral. It seems complicated since Wolfram cannot generate the steps for this integral.
Any suggestion is highly appreciated.

I think that the statement contains a misprint. Verify if it is not $x^{2}-%
\frac{1}{8}\ln x$ instead. If it is the case then
\begin{eqnarray*}
1+\left( f^{\prime }(x)\right) ^{2} &=&1+(2x-\frac{1}{8x})^{2} \\
&=&1+(2x)^{2}-2(2x)(\frac{1}{8x})+(\frac{1}{8x})^{2} \\
&=&1+(2x)^{2}-\frac{1}{2}+(\frac{1}{x})^{2} \\
&=&(2x)^{2}+(1-\frac{1}{2})+(\frac{1}{x})^{2} \\
&=&(2x)^{2}+\frac{1}{2}+(\frac{1}{x})^{2} \\
trick\ here &\rightarrow &=(2x)^{2}+2(2x)(\frac{1}{8x})+(\frac{1}{8x})^{2} \\
&=&(2x+\frac{1}{8x})^{2}.
\end{eqnarray*}
EDIT:
Assume that $f(x)$ is $ax^{2}-b\ln x,$ and ask whether $1+(f^{\prime
}(x))^{2}$ is a perfect square.
\begin{eqnarray*}
1+\left( f^{\prime }(x)\right) ^{2} &=&1+(2ax-\frac{b}{x})^{2} \\
&=&1+(2ax)^{2}-2(2ax)(\frac{b}{x})+(\frac{b}{x})^{2} \\
&=&(2ax)^{2}+(1-4ab)+(\frac{b}{x})^{2}
\end{eqnarray*}
this is a perfect square if 
\begin{equation*}
1-4ab=4ab
\end{equation*}
and in this case 
\begin{eqnarray*}
1-4ab &=&4ab \\
&=&2(2ax)(\frac{b}{x})
\end{eqnarray*}
then
\begin{eqnarray*}
1+\left( f^{\prime }(x)\right) ^{2} &=&(2ax)^{2}+2(2ax)(\frac{b}{x})+(\frac{b%
}{x})^{2} \\
&=&(2ax+\frac{b}{x})^{2}.
\end{eqnarray*}
So $ab=\frac{1}{8}.$ That is why, probably for your case $a=1$ then $b$
should be $\frac{1}{8}.$ But it is also possible that $ab=\frac{1}{8}$
for other combinations. $(a=\frac{1}{2},\ b=\frac{1}{4},\ ....)$
If you browse some textbook (for calculus courses), you will find that each
related exercise is constructed in this way. I mean each time $(1+\left(
f^{\prime }(x)\right) ^{2})$ is a perfect square!
