How could one solve $\int_{0}^{\infty} \frac{1}{1-t^4}dt$ with special functions?

(Assuming principal value)

Let $t=u^4$ as you suggested, then we get $$ PV\int_{0}^{+\infty}\frac{du}{4u^{3/4}(1-u)}= \lim_{\varepsilon\to0}\left[\int_{0}^{1-\varepsilon}\frac{du}{4u^{3/4}(1-u)}+\int_{1+\varepsilon}^{+\infty}\frac{du}{4u^{3/4}(1-u)}\right] $$ and now an idea could be letting $u=1/(1-y)$ in the second integral to get $$ -\frac{1}{4}\int_{\varepsilon/(1+\varepsilon)}^{1}y^{-1}(1-y)^{-1/4}dy, $$ so $$ PV\int_{0}^{+\infty}\frac{du}{4u^{3/4}(1-u)}=\frac{1}{4} \lim_{\varepsilon\to0}\left[\int_{0}^{1-\varepsilon}u^{-3/4}(1-u)^{-1}du-\int_{\varepsilon/(1+\varepsilon)}^{1}y^{-1}(1-y)^{-1/4}dy\right]. $$ Now $$ \beta(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}, $$ and since the original integral is not well defined, we get $+\infty-\infty$. Letting also $u=1-y$ in the first term, \begin{align} PV\int_{0}^{+\infty}\frac{dt}{1-t^4}&=\frac{1}{4} \lim_{\varepsilon\to0}\left[\int_{\varepsilon}^{1}(1-y)^{-3/4}y^{-1}dy-\int_{\varepsilon/(1+\varepsilon)}^{1}y^{-1}(1-y)^{-1/4}dy\right]\\ &=\frac{1}{4} \int_{0}^{1}\left[(1-y)^{-3/4}-(1-y)^{-1/4}\right]y^{-1}dy\\ &=\frac{1}{4} \int_{0}^{1}\frac{u^{-3/4}-u^{-1/4}}{1-u}du=\frac{1}{2}\int_{0}^{1}\frac{1-\sqrt u}{u^{1/4}(1-u)}\frac{du}{2\sqrt u}\\ &=\frac{1}{2}\int_{0}^{1}s^{-1/2}(1+s)^{-1}ds\\ &=\left[\tan^{-1}(\sqrt s)\right]_0^1=\pi/4, \end{align} where in the second step $$ \left| \int_{\varepsilon/(\varepsilon+1)}^\varepsilon y^{-1}(1-y)^{-1/4}dy\right|\le \frac{\varepsilon +1}{\varepsilon}\int_{\varepsilon/(\varepsilon+1)}^\varepsilon dy=\frac{\varepsilon+1}{\varepsilon}\frac{\varepsilon^2}{\varepsilon+1}\to0 $$ has been used.

If you admit complex methods in general, let $$ PV\int_0^{+\infty}\frac{dt}{1-t^4}=\frac{1}{2}PV\int_{-\infty}^{+\infty}\frac{dt}{1-t^4}. $$ Now, using a half-circle $C$ in the upper half complex plane with indentations at the poles, we have a residue coming from the pole in the upper-half complex plane $$ \oint_{C}\frac{dz}{1-z^4}=i2\pi\frac{1}{4i}=\frac{\pi}{2} $$ expanding the integration contour, neglecting the contribution from the large circle and evaluating the contribution from the poles on the real line, which in fact vanishes, $$ PV\int_{-\infty}^{+\infty}\frac{dt}{1-t^4}-i\pi\frac{1}{3}+i\pi\frac{1}{3}=\frac{\pi}{2}. $$ So $$ PV\int_{0}^{+\infty}\frac{dt}{1-t^4}=\frac{\pi}{4}, $$