Here is an idea if you really don't want to use residues. Consider $$f(k)=\int_{-\infty}^{\infty}\frac{\cos(kx)}{x^4+1}dx.$$
Then, consider the fourth derivative of $f$ as a distribution:
$$f''''(k)=\int_{-\infty}^\infty\frac{x^4\cos(kx)}{x^4+1}dx.$$
That is,
$$f''''(k)+f(k)=\int_{-\infty}^\infty\cos(kx)dx=\int_{-\infty}^\infty e^{ikx}dx.$$
Since you are taking physics, you probably know that
$$\int_{-\infty}^\infty e^{ikx}dx=2\pi\delta(k).$$
Hence,
$$f''''(k)+f(k)=2\pi\delta(k).$$
Therefore, we expect $f'''(k)$ to jump with step size $2\pi$ at $k=0$. Knowing $$f'''(k)=\int_{-\infty}^{\infty}\frac{x^3\sin(kx)}{x^4+1}dx,$$
we have $f'''(0)=0$, and since $f'''$ is an odd function, we have
$$\lim_{\epsilon\searrow0}f'''(\epsilon)=\pi$$
and
$$\lim_{\epsilon\searrow0}f'''(-\epsilon)=-\pi.$$
We also find that
$$f''(k)=-\int_{-\infty}^\infty\frac{x^2\cos(kx)}{x^4+1}dx$$
so
$$f''(0)=-\frac{\pi}{\sqrt{2}}$$
by any means you know (partial fractions, etc). We also have $f'(0)=0$ (this is trivial) and $f(0)=\frac{\pi}{\sqrt{2}}$ (this requires some work).
Now, from a standard differential equation course, you should be able to see that the homogeneous solution $g''''(k)+g(k)=0$ is of the form $$g(k)=e^{\frac{k}{\sqrt{2}}}\Biggl(p\cos\left(\frac{k}{\sqrt{2}}\right)+q\sin\left(\frac{k}{\sqrt{2}}\right)\Biggr)+e^{-\frac{k}{\sqrt{2}}}\Biggl(r\cos\left(\frac{k}{\sqrt{2}}\right)+s\sin\left(\frac{k}{\sqrt{2}}\right)\Biggr).$$ So, our answer $f$ should be of the form
$$f(k)=\begin{cases}g(k)&\text{if }k>0,\\\frac{\pi}{\sqrt{2}}&\text{if }k=0,\\g(-k)&\text{if }k<0,\end{cases}$$
where $g$ is a homogeneous solution satisfying
$$g(0)=\frac{\pi}{\sqrt{2}},\ g'(0)=0,\ g''(0)=-\frac{\pi}{\sqrt{2}},\ \wedge\ g'''(0)=\pi.$$
(We can also just write $f(k)=g\big(|k|\big)$.)
This gives
$$g(k)=\pi e^{-\frac{k}{\sqrt{2}}}\left(\frac{\cos\left(\frac{k}{\sqrt{2}}\right)+\sin\left(\frac{k}{\sqrt{2}}\right)}{\sqrt{2}}\right).$$
(You can cheat here a bit. Since $f(k)$ is bounded, the coefficients $p$ and $q$ have to be $0$. So, you only need two equations to solve for $r$ and $s$. And if you used only $\frac{s-r}{\sqrt{2}}=g'(0)=f'(0)=0$ and $\frac{r+s}{\sqrt{2}}=g'''(0)=\pi$ to solve for $r$ and $s$, there would be no need to compute cumbersome integrals to get $g(0)=f(0)=\frac{\pi}{\sqrt{2}}$ and $g''(0)=f''(0)=-\frac{\pi}{\sqrt{2}}$.)
In other words,
$$\int_{-\infty}^{\infty}\frac{\cos(kx)}{x^4+1}dx=f(k)=\pi e^{-\frac{|k|}{\sqrt{2}}}\left(\frac{\cos\left(\frac{|k|}{\sqrt{2}}\right)+\sin\left(\frac{|k|}{\sqrt{2}}\right)}{\sqrt{2}}\right).$$
In particular, when $k=1$, we have
$$\int_{-\infty}^{\infty}\frac{\cos(kx)}{x^4+1}dx=\pi e^{-\frac{1}{\sqrt{2}}}\left(\frac{\cos\left(\frac{1}{\sqrt{2}}\right)+\sin\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}}\right)\approx 1.54428.$$
But frankly, I feel like using residues is the most straightforward direction, and possibly leads to less tedious computations.
