<s>[INST] Evaluation of the integral $\int \sqrt{t^4-t^2 + 1}\,dt$ My friend took his Calculus $2/3$ test yesterday.
One of the questions he had trouble with was this integral:
$$\int \sqrt{t^4-t^2 + 1}dt$$
My attempt
It seems rather clear that the only approach was trigonometric substitution.
First, completing the square:
$$\int \sqrt{t^4-t^2 + 1}dt = \int\sqrt{t^4-t^2 + \frac{1}{4} + \frac{3}{4}}dt = \int \sqrt{\left(t^2 - \frac{1}{2}\right) + \frac{3}{4}}dt$$
Next, I let $$\sec \theta = \frac{\sqrt{\left(t^2 - \frac{1}{2}\right) + \frac{3}{4}}}{\frac{\sqrt{3}}{2}}$$
$$\frac{\sqrt{3}}{2}\sec \theta = \sqrt{\left(t^2 - \frac{1}{2}\right) + \frac{3}{4}}$$
$$\tan \theta = \frac{t^2 - \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{2t^2 - 1}{\sqrt{3}}$$
$$\sec^2 \theta d\theta = \frac{4}{\sqrt{3}}tdt,\ dt = \frac{\sqrt{3}\sec^2 \theta}{4t} d\theta$$
$$t = \sqrt{\frac{1}{2}\left(\sqrt{3}\tan \theta + 1\right)}$$

Substituting this all in:
$$\int \left(\frac{\sqrt{3}}{2} \sec \theta\right) \frac{\sqrt{3}\sec^2 \theta}{4\sqrt{\frac{1}{2}\left(\sqrt{3}\tan \theta + 1\right)}} d\theta$$
How would I approach this from here?
I'm thinking of using u-substitution but I'm sure that it would bring me back to where I started, meaning I would have to use trig. substitution again.
 [/INST] The indefinite integral cannot be expressed in terms of elementary functions. Evaluating it requires knowledge of elliptic integrals. However, it is interesting to note that its definite counterpart, when evaluated over the entire real line, and subtracted from its quadratic or parabolic asymptote, yields 
$$\int_{-\infty}^\infty\bigg[\sqrt{t^4-t^2+1}-\bigg(t^2-\dfrac12\bigg)\bigg]dt~=~\dfrac23K\bigg(\dfrac34\bigg)+\dfrac43E\bigg(\dfrac34\bigg),$$
which is nothing else than the volume of the oloid, whose decimal expansion can be found on OEIS
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