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.
We have:

$$I=\int \sqrt{4x^2-3+x^{-2}}dx$$

doing substitution $t=x^2$ we have $dx=\frac{1}{2\sqrt{t}}dt$ and

$$\int \frac{\sqrt{4t-3+\frac{1}{t}}}{2\sqrt{t}}dt=\frac 1 2 \int \sqrt{4-\frac 3 t + \frac 1 {t^2}}dt=\frac 1 2 \int \frac 1 t \sqrt{4t^2-3t+1}dt=\frac 1 2 \int \frac{4t^2-3t+1}{t\sqrt{ 4t^2-3t+1}}dt$$

using the equality $4t^2-3t+1=\frac t 2 (8t-3) -\frac 3 2 t +1$:

$$\frac 1 4 \int \frac{8t-3}{\sqrt{4t^2-3t+1}}dt+\frac 1 4 \int \frac{2-3t}{t\sqrt{4t^2-3t+1}}dt=\frac 1 2 \sqrt{4t^2-3t+1} +\underbrace{\frac 1 2 \int \frac 1 {t\sqrt{4t^2-3t+1}}dt}_1 - \underbrace{\frac 3 4 \int \frac 1 {\sqrt{4t^2-3t+1}} dt}_2$$

Integral #1

$$\frac 1 2 \int \frac {dt}{t\sqrt{4t^2-3t+1}}$$

Doing substitution $u=\frac 1 t$, $du = -\frac{1}{t^2}dt$,

$$-\frac 1 2 \int \frac {du}{u\sqrt{\frac 4 {u^2} - \frac 3 u + 1}}=-\frac 1 2 \int \frac {du}{\sqrt{u^2-3u+4}}=-\frac 1 2 \int \frac{du}{\sqrt{(u-\frac 3 2)^2+\frac 7 4}}$$

Substitution $s=u-\frac 3 2$, $ds=du$:

$$-\frac 1 2 \int \frac{ds}{\sqrt{s^2+\frac 7 4}}=-\frac 1 2 \ln\left| s+\sqrt{s^2+\frac 7 4} \right|+C_1=-\frac 1 2 \ln\left| u-\frac 3 2+\sqrt{u^2-3u+4} \right|+C_1=-\frac 1 2 \ln\left| \frac 1 t-\frac 3 2+\sqrt{\frac{1}{t^2}-\frac 3 t+4} \right|+C_1=-\frac 1 2 \ln\left| \frac 1 {x^2}-\frac 3 2+\sqrt{\frac{1}{x^4}-\frac 3 {x^2}+4} \right|+C_1$$

Integral #2

$$\frac 3 4 \int \frac 1 {\sqrt{4t^2-3t+1}} dt=\frac 3 8 \int \frac 1 {\sqrt{t^2-\frac 3 4 t+\frac 1 4}} dt=\frac 3 8 \ln\left|t-\frac 3 8 + \sqrt{t^2-\frac 3 4 t+\frac 1 4}\right|+C_2=\frac 3 8 \ln\left|x^2-\frac 3 8 + \sqrt{x^4-\frac 3 4 x^2+\frac 1 4}\right|+C_2$$

Summary

$$I=\frac 1 2 \sqrt{4x^4-3x^2+1} -\frac 1 2 \ln\left| \frac 1 {x^2}-\frac 3 2+\sqrt{\frac{1}{x^4}-\frac 3 {x^2}+4} \right|  - \frac 3 8 \ln\left|x^2-\frac 3 8 + \sqrt{x^4-\frac 3 4 x^2+\frac 1 4}\right|+C$$

Calculating exact value for bounds $[1;e]$:

$$[I]_1^e=\frac 1 2 \sqrt{4e^4-3e^2+1} -\frac 1 2 \ln\left| \frac 1 {e^2}-\frac 3 2+\sqrt{\frac{1}{e^4}-\frac 3 {e^2}+4} \right|  - \frac 3 8 \ln\left|e^2-\frac 3 8 + \sqrt{e^4-\frac 3 4 e^2+\frac 1 4}\right|-
\frac 1 2 \sqrt{4-3+1} +\frac 1 2 \ln\left| 1-\frac 3 2+\sqrt{1-3+4} \right|  + \frac 3 8 \ln\left|1-\frac 3 8 + \sqrt{1-\frac 3 4+\frac 1 4}\right|
\\=\frac 1 2 \sqrt{4e^4-3e^2+1} -\frac 1 2 \ln\left| \frac{1+\frac{3}{2}e^2+\sqrt{1-3 e^2+4e^4}}{e^2}\right|  - \frac 3 8 \ln\left|\frac{8e^2-3 + 4\sqrt{4e^4-3 e^2+1}}{8}\right|-
\frac{\sqrt{2}}{2} +\frac 1 2 \ln\left| -\frac 1 2+\sqrt{2} \right|  + \frac 3 8 \ln\left|\frac{13}{8}\right|$$

Note

In my calculation I used Euler's substitution to get the following formula:

$$\int \frac{dx}{\sqrt{x^2+A}} = \ln\left|x+\sqrt{x^2+A}\right|+C$$
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} &amp;=&amp;1+(2x-\frac{1}{8x})^{2} \\
&amp;=&amp;1+(2x)^{2}-2(2x)(\frac{1}{8x})+(\frac{1}{8x})^{2} \\
&amp;=&amp;1+(2x)^{2}-\frac{1}{2}+(\frac{1}{x})^{2} \\
&amp;=&amp;(2x)^{2}+(1-\frac{1}{2})+(\frac{1}{x})^{2} \\
&amp;=&amp;(2x)^{2}+\frac{1}{2}+(\frac{1}{x})^{2} \\
trick\ here &amp;\rightarrow &amp;=(2x)^{2}+2(2x)(\frac{1}{8x})+(\frac{1}{8x})^{2} \\
&amp;=&amp;(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} &amp;=&amp;1+(2ax-\frac{b}{x})^{2} \\
&amp;=&amp;1+(2ax)^{2}-2(2ax)(\frac{b}{x})+(\frac{b}{x})^{2} \\
&amp;=&amp;(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 &amp;=&amp;4ab \\
&amp;=&amp;2(2ax)(\frac{b}{x})
\end{eqnarray*}
then
\begin{eqnarray*}
1+\left( f^{\prime }(x)\right) ^{2} &amp;=&amp;(2ax)^{2}+2(2ax)(\frac{b}{x})+(\frac{b%
}{x})^{2} \\
&amp;=&amp;(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!