Inverse Fourier Transform I've got a problem where I need to find the IFT of
$$F(\omega) = \frac{1 + i\omega}{6-\omega^2+5i\omega}$$
I've been trying to solve it through partial fractions, but that gives us 
$$\frac{1 + i\omega}{6-\omega^2+5i\omega} = \frac{1 + i\omega}{(-\omega + 3i)(\omega-2i)} = \frac{i(\omega - i)}{(-\omega+3i)(\omega-2i)}= \frac{A(\omega-2i) + B(-\omega+3i)}{(-\omega + 3i)(\omega-2i)} $$
and I'm not entirely certain how to solve for A and B here.
Substituting it into the IFT formula give us
$$\frac 1{2\pi} \int_{-\infty}^\infty \frac{1 + i\omega}{6-\omega^2+5i\omega} \cdot e^{i\omega t} \ d\omega$$
or any combination of the above factorizations, but I have no idea how to even begin integrating something like this.
Anything, even a hint at what I should do in such a situation like this would help. 
Thanks a lot for taking a look!
 A: By using partial fractions we have
\begin{align}
\frac{i\omega +1}{6-\omega^2+5i\omega}&=\frac{i\omega+1}{(i\omega+2)(i\omega+3)}\\
&=\frac{A}{i\omega+3}+\frac{B}{i\omega+2}
\end{align}
Where $A$ and $B$ are constants such that 
\begin{align}
A(i\omega+2)+B(i\omega+3)&=i\omega+1\\
(A+B)i\omega+2A+3B&=i\omega+1
\end{align}
Then, solving the linear system of equations
\begin{align}
A+B&=1\\
2A+3B&=1
\end{align}
we get $\color{red}{A=2}$ and $\color{red}{B=-1}$.
Hence
\begin{align}
\frac{i\omega +1}{6-\omega^2+5i\omega}&=\frac{2}{i\omega+3}-\frac{1}{i\omega+2}
\end{align}
From here you can apply the inverse FT:
\begin{align}
\mathscr{F}^{-1}\left\{\frac{i\omega +1}{6-\omega^2+5i\omega}\right\}&=2\mathscr{F}^{-1}\left\{\frac{1}{i\omega+3}\right\}-\mathscr{F}^{-1}\left\{\frac{1}{i\omega+2}\right\}\\[5pt]
&=\color{blue}{\boxed{\left(2e^{-3t}-e^{-2t}\right)u(t)}}
\end{align}
A: We can evaluate the integral using contour integration.  First, let 
$$F(\omega) = \frac{1 + i\omega}{6-\omega^2+5i\omega}
$$
Then, the inverse Fourier Transform $\mathscr{F}^{-1}\{F\}(t)$ is given by 
$$\begin{align}
\mathscr{F}^{-1}\{F\}(t)&=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)e^{i\omega t}\,d\omega\\\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1 + i\omega}{6-\omega^2+5i\omega}e^{i\omega t}\,d\omega\\\\
&=-\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1 + i\omega}{(\omega-3i)(\omega-2i)}e^{i\omega t}\,d\omega\\\\
\end{align}$$
We now analyze the closed-contour integral $I(t)$ as given by
$$I(t)=-\frac1{2\pi}\oint_C \frac{1 + iz}{(z-3i)(z-2i)}e^{iz t}\,dz$$
where $C$ is the contour comprised of the line segment along the real axis from $-R$ to $R$ and a semi-circular arc $C_R$, with radius R, centered at the origin in the upper-half plane (lower-half plane) for $t>0$ ($t<0$).  
Note that the integrand has poles at $z =3i$ and $z=2i$.  Therefore, from the residue theorem we have
$$I(t)=   
\begin{cases}
2e^{-3t}-e^{-2t}&, t>0\\\\\
0&,t<0
\end{cases}$$
Then, we have
$$\begin{align}I(t)&=-\frac1{2\pi}\int_{-R}^R\frac{1 + i\omega}{(\omega-3i)(\omega-2i)}e^{i\omega t}\,d\omega+\int_0^{\pm \pi}\frac{1 + iRe^{i\phi}}{(Re^{i\phi}-3i)(Re^{i\phi}-2i)}e^{iRe^{i\phi} t}\,iRe^{i\phi}\,d\phi\\\\&=\begin{cases}
2e^{-3t}-e^{-2t}&, t>0 \tag 1\\\\\
0&,t<0
\end{cases}
\end{align}$$
The first integral on the right-hand side of $(1)$ approaches $\mathscr{F}^{-1}\{F\}(t)$ as $R\to \infty$, while the second integral on the right-hand side of $(1)$ goes to $0$.  
Putting it all together, we arrive at 
$$\mathscr{F}^{-1}\{F\}(t)\begin{cases}
2e^{-3t}-e^{-2t}&, t>0 \\\\\
0&,t<0
\end{cases}$$
