Problems involving the derivations of CTFT $X(j\omega)$ for a given signal $x(t)$ or the reverse: finding $x(t)$ for a given $X(j\omega)$ can be solved in a number of ways. Fundamentally there exists the so called fourier integrals: the Analysis $$X(j\omega) = \int_{-\infty}^{\infty} {x(t) e^{-j\omega t} dt }$$
and the synthesis: $$x(t) = \frac {1}{2\pi} \int_{-\infty}^{\infty} {X(j\omega) e^{j\omega t} d\omega }$$
to directly compute what is being asked. But due to several reasons we also develop practical, shortway techniques to find those signals, as a very natural extension of engineering practice.
One such practical method of finding the forward or inverse fourier transforms involves the systematic application of known pairs with partial fraction expansions. Which enables the bypass of the integrals and yields fast answers.
First of all you shall apply the method of partial fraction expansion to your given CTFT $$X(j\omega) = \frac {3}{(1+j\omega)(2-j\omega)} = \frac {1}{1+j\omega} + \frac {1}{2-j\omega} $$
Then you shall consult into a table of fourier pairs, which would guide you to find which signal has transforms of this kind. From those standard tables one can see that $x(t) = e^{-at} u(t) $ has the transform of $1/(a + j\omega)$ and also $x(t) = e^{at} u(-t) $ has the transform of $1/(a - j\omega)$, for $ a> 0$
Henceforth, you would conlcude by finally invoking the linearity property of CTFT that $$ x(t) = e^{-t} u(t) + e^{2t} u(-t)$$
