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.

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.
