Technique for Fourier transform I was looking at this exercise

Evaluate the Fourier transform of the following function $$f(x)=\frac{xe^{5ix}}{x^2+2x+10}$$

and the professor was using some techniques which I don't quite understand. I think that has something to do with the properties for translation and scaling for the Fourier transform, but I cannot work it out. 
First of all, we define the Fourier transform in the following manner $$\hat{f}(k)={1\over {\sqrt{2\pi}}}\int_{\mathbb{R}}e^{ikx}f(x)dx$$ 
Then the steps that the professor takes are as follows:
$$\begin{align}
&&\frac{1}{a^2+x^2}&&\overset{\mathcal{F}}{\longrightarrow} &&\frac{\pi}{a}e^{-a|k|}\tag1 \\
&[x\rightarrow x+1]&\frac{1}{x^2+2x+1+a^2}&&\overset{\mathcal{F}}{\longrightarrow} &&\frac{\pi}{a}e^{-a|k|+ik}\tag2 \\
&[a=3]&\frac{1}{x^2+2x+10}&&\overset{\mathcal{F}}{\longrightarrow} &&\frac{\pi}{3}e^{-3|k|+ik}\tag3 \\
&[f(x)\rightarrow xf(x)]&\frac{x}{x^2+2x+10}&&\overset{\mathcal{F}}{\longrightarrow} &&\frac{\pi}{3}e^{-3|k|+ik}(1+3i\operatorname{sgn}(k))\tag4\\
&[f(x)\rightarrow e^{5ix}f(x)]&\frac{xe^{5ix}}{x^2+2x+10}&&\overset{\mathcal{F}}{\longrightarrow} &&\frac{\pi}{3}e^{-3|k-5|+i(k-5)}(1+3i\operatorname{sgn}(k-5))\tag5
\end{align}$$ 
From what I undestrand the second step takes advantage of the translation property for the Fourier transform. The second step is trivial (clever, I know). The problems start at the last two steps: my thought was that for the fourth step he's using the convolution theorem, mainly 

$$\mathcal{F}\{f\cdot g\} = \mathcal{F}\{f\}*\mathcal{F}\{g\}$$

and easily the Fourier transform of $x$ is just a Dirac delta, so the convolution is trivial. On the last step I don't have any clue on what property he's using! 
Could someone explain to me what is the professor doing and if my thoughts are right or not? 
 A: For the the second-to-last step, he's using the property that F[xf] = (F[f])'/i; that is, multiplying by x and taking the Fourier Transform is the same as taking the Fourier Transform, taking the derivative of that, and then dividing by i (or, equivalently, multiplying by -i). For the last step, he's simply using the definition of the Fourier Transform: to take the Transform, you integrate $e^{ikx}f(x)$. So to get the Transform of $e^{5ix}f(x)$, you integrate $e^{ikx}e^{5ix}f(x)=e^{ikx+5ix}f(x)=e^{i(k+5)x}f(x)$. If we define $u=k+5$, then this is just the Fourier Transform of $f(x)$, except that the transformed function is a function of $u$. To get back to $k$, we subtract 5. So your professor just replaced all the instances of $k$ in the the second-to-last step with $k-5$. That is, the "k" in the second-to-last step is 5 less than the "k" in the last step.
If you're having trouble with that:
Take
$\int e^{i(t+5)x}f(x)$
set $u = t+5$
$\int e^{iux}f(x)=F[f](u)$
since $u=t+5$, $t=u-5$, so $F[f](u)=F[f](t-5)$
So we take the Fourier Transform of $f$, then replace all instances of $k$ with $k-5$.
Remember, the Fourier Transform $F[f](k)$ is simply the strength of the $e^{ikx}$ component of $f$. By multiplying $f$ by $e^{5ix}$, we're shifting all the components by 5: the $k=0$ component becomes the $k=5$ component, etc. 
There's a dual aspect to the translation property of Fourier Transforms: translating $f$ by $h$ multiplies the transform by $e^{ihx}$, and multiplying $f$ by $e^{ihx}$ translates the transform by $-h$.
