Using the complex representation makes the problem solvable with a first order ODE.
$$I(a)=\int_0^\infty\frac{e^{iax^2}}{x^2+1}dx,$$
$$I'(a)=i\int_0^\infty x^2\frac{e^{iax^2}}{x^2+1}dx,$$
$$I'(a)+iI(a)=i\int_0^\infty e^{iax^2}dx=wa^{-1/2},$$ where $w$ is a complex constant (namely $(i-1)\sqrt{\pi/8}$).
Then by means of an integrating factor,
$$(I'(a)+iI(a))e^{ia}=(I(a)e^{ia})'=wa^{-1/2}e^{ia}$$
and integrating from $a=0$ to $1$,
$$I(a)e^{ia}-I(0)=w\int_0^1a^{-1/2}e^{ia}da=2w\int_0^1e^{ib^2}db=2w(C(1)+iS(1)).$$
Finally,
$$I(a)=\left(2w(C(1)+iS(1))+I(0)\right)e^{-i}$$ of which you take the real part. (With $I(0)=\pi/2$.)
Note that we have been using the Fresnel integral without the $\pi/2$ factor in its definition, and this answer coincides with those of the CAS softwares.
Clearly $z_1=\frac{\sqrt2}2(1+i)$ is the only root of $z^4+1=0$ in the first quadrant.
For $R>1$, defining $\alpha_R(t)=t, \beta_R(t)=it$ for $t \in [0,R]$ and finally $\gamma_R(t)=Re^{i\theta}$ for $\theta\in[0,\frac\pi2]$. Then
$$ \frac{1}{2\pi i}\bigg(\int_{\alpha_R}+\int_{\gamma_R}-\int_{\beta_R}\bigg)\frac{e^z}{z^4+1}dz=\frac{e^{iz_1}}{3z_1}$$
or
$$ \int_0^R\frac{e^{it}}{t^4+1}dt+\int_0^{\pi/2}\frac{e^{iRe^{4i\theta}}}{R^4e^{4i\theta}+1}iRe^{i\theta}d\theta-\int_0^R\frac{e^{-t}}{t^4+1}idt=2\pi i\frac{e^{iz_1}}{3z_1} $$
Letting $R\to\infty$ gives
$$ \int_0^\infty\frac{e^{it}}{t^4+1}dt-\int_0^\infty\frac{e^{-t}}{t^4+1}idt=2\pi i\frac{e^{iz_1}}{3z_1} $$
or
$$ \int_0^\infty\frac{e^{it}}{t^4+1}dt=i\int_0^\infty\frac{e^{-t}}{t^4+1}dt+2\pi i\frac{e^{iz_1}}{3z_1}. \tag1 $$
and so on and on.
Update:
\begin{eqnarray}
\int_0^\infty\frac{e^{-t}}{t^4+1}dt&=&\int_0^1\frac{e^{-t}}{t^4+1}dt+\int_1^\infty\frac{e^{-t}}{t^4+1}dt\\
&=&\int_0^1\sum_{n=0}^\infty (-1)^n t^{4n}e^{-t}dt+\int_1^\infty\sum_{n=0}^\infty (-1)^n t^{-4n-4}e^{-t}dt\\
&=&\sum_{n=0}^\infty (-1)^n\bigg[\Gamma (4 n+1)+\Gamma (-4 n-3,1)-\Gamma (4 n+1,1)\bigg].
\end{eqnarray}
which seems hard to simplify.
Best Answer
Without complex analysis or residues, you can compute the antiderivative.
Use the roots of unity and then partial fraction decomposition to face four integrals looking like $$I_a=\int \frac {e^{-t}}{t+a}$$ where $a$ is a complex number. This makes $$I_a=e^a\, \text{Ei}(-(t+a))$$
For sure, the final result is not very pretty $$\int_0^{+\infty} \frac{e^{-t}}{t^4+1}dt=\frac{1}{4 \sqrt{2} \pi ^{3/2}}\,\,G_{1,5}^{5,1}\left(\frac{1}{256}\right.\left| \begin{array}{c} \frac{3}{4} \\ 0,\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{3}{4} \end{array} \right)$$ where appears the Meijer G function.
Numerically, this is $0.63047783491849835735$ which is not recognized by inverse symbolic calculators.