$$f(x) = \frac{\cos(bx)}{a^2 +x^2} $$ for $a, b$ $>0$
$$\hat f(\omega) = \int\limits^{\infty}_{-\infty} \dfrac{1}{2{\pi}}\cdot\dfrac{\cos\left(bx\right)}{a^2+x^2}\mathrm{e}^{-\mathrm{i}wx}\,\mathrm{d}x = \int\limits^{\infty}_{-\infty} \dfrac{\mathrm{e}^{\mathrm{i}bx}+\mathrm{e}^{-\mathrm{i}bx}}{4\pi\left(a^2+x^2\right)}\mathrm{e}^{-\mathrm{i}wx}\,\mathrm{d}x=$$
$$ \int\limits^{\infty}_{-\infty} \dfrac{\mathrm{e}^{\mathrm{i}x\left(b-w\right)}}{2\left(a^2+x^2\right)}\,\mathrm{d}x + \int\limits^{\infty}_{-\infty} \dfrac{\mathrm{e}^{-\mathrm{i}x\left(b+w\right)}}{2\left(a^2+x^2\right)}\,\mathrm{d}x$$
How should I continue from here?
I checked on wolfram alpha but the answer should not expressed with dirac function since we didn't learn it.
Note that the factor $\cos(bx)$ is just a modulation, i.e., a translation of the spectrum. So we can try to determine the Fourier transform $G(\omega)$ of
$$g(x)=\frac{1}{a^2+x^2}\tag{1}$$
and from $G(\omega)$ determine the Fourier transform of $f(x)$:
$$F(\omega)=\frac12\big[G(\omega-b)+G(\omega+b)\big]\tag{2}$$
Rewrite $g(x)$ as
$$g(x)=\frac{1}{2a}\left[\frac{1}{a+jt}+\frac{1}{a-jt}\right]\tag{3}$$
It's straightforward to see that
$$\mathcal{F}\left\{\frac{1}{a+jt}\right\}=2\pi e^{a\omega}u(-\omega),\qquad a>0\tag{4}$$
and
$$\mathcal{F}\left\{\frac{1}{a-jt}\right\}=2\pi e^{-a\omega}u(\omega),\qquad a>0\tag{5}$$
where $u(\omega)$ is the Heaviside unit step function.
Using $(4)$ and $(5)$, the Fourier transform of $g(x)$ is obtained as
$$G(\omega)=\frac{\pi}{a}e^{-a|\omega|}\tag{6}$$
The Fourier transform of $f(x)$ can be obtained by plugging $(6)$ into $(2)$.