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.
 A: After substituting $t=(8x^2-3)/\sqrt{7}$, we get
$$\int\sqrt{4x^2-3+\frac{1}{x^2}}dx=\frac{7}{8}\int\frac{\sqrt{t^2+1}}{\sqrt{7}t+3}dt.$$
Making another substitution, $t=\sinh s$, we obtain
$$\frac{7}{8}\int\frac{\sqrt{t^2+1}}{\sqrt{7}t+3}dt=
\frac{\sqrt{7}}{8}\int\frac{\sinh^2 s+1}{\sinh s+\displaystyle\frac{3}{\sqrt{7}}}ds=\frac{\sqrt{7}}{8}\int\sinh s~ds-\frac{3}{8}\int ds+\frac{2}{\sqrt{7}}\int\frac{1}{\sinh s+\displaystyle\frac{3}{\sqrt{7}}}ds.$$
The first two terms are trivially integrable; as to the third term, I cheated and checked my Gradshteyn ($\int (a+b\sinh x)^{-1}dx=2(a^2+b^2)^{-1/2}{\rm atanh}\big([a\tanh(x/2)-b][a^2+b^2]^{-1/2}\big)$). After simplification, we get
$$\int\sqrt{4x^2-3+\frac{1}{x^2}}dx=\frac{1}{2}\sqrt{4x^4-3x^2+1}-\frac{3}{8}{\rm asinh}~\frac{8x^2-3}{\sqrt{7}}+{\rm atanh}\left(\frac{3-\sqrt{7}}{4}-\frac{3\sqrt{7}}{16x^2-6+2\sqrt{7}+8\sqrt{4x^4-3x^2+1}}\right).$$
Up to an integration constant, this agrees with the result given by Maple.
A: 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!
