This is an integral that has been giving me trouble in my Complex Analysis course. I have already learned residues in one of my physics courses, however, we are not allowed to use it since we have not yet covered it in this course.
I am honestly pretty stumped on this one. I'm assuming that since this is a real integral, we should probably break the denominator into (x^2 + i)*(x^2-i) to turn it into a complex integral and maybe evaluate using the Cauchy integral formula, but I'm having a hard time convincing myself that this is the right approach. Any and all insight is much appreciated.
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.