Evaluate integral: $$ I_1=\int\limits_{-\infty}^{+\infty}\frac{\cos 3x}{5+6x+9x^2}dx $$
I used Fourier transform in the following form: $$ \hat{f}(y)=F[f(x)]=\int\limits_{-\infty}^{+\infty}f(x)e^{ixy}dx $$ Therefore, I have the following properties and formulas: $$ \begin{aligned} &F[f(kx)]=\frac{1}{|k|}\hat{f}\left(\frac{y}{k}\right),\ \ k\ne 0\\ &F[f(x-a)]=e^{iay}\hat{f}(y),\ \ a\in\mathbb{R}\\ &F\left[\frac{1}{1+x^2}\right]=\pi\cdot e^{-|y|} \end{aligned} $$
So, here is my attempt:
Suppose $I_2=i\cdot\int\limits_{-\infty}^{+\infty}\frac{\sin 3x}{5+6x+9x^2}dx$. Therefore: $$ \begin{aligned} &I_1+I_2=\int\limits_{-\infty}^{+\infty}\frac{e^{3ix}}{5+6x+9x^2}dx=F\left[\frac{1}{5+6x+9x^2}\right]\Bigm|_{y=3}\\ &\frac{1}{5+6x+9x^2}=\frac{1}{4}\cdot\frac{1}{\left(\frac{3x}{2}+\frac{1}{2}\right)^2+1}=g(x)\\ &F\left[\frac{1}{\left(\frac{3x}{2}\right)^2 +1}\right]=\frac{2}{3}\pi e^{-\frac{2}{3}|y|}\ \ \Rightarrow\ \ F\left[\frac{1}{\left(\frac{3x}{2}-\left(-\frac{1}{2}\right)\right)^2 +1}\right]=e^{-\frac{1}{2}iy}\cdot\frac{2}{3}\pi e^{-\frac{2}{3}|y|}\\ &F[g(x)]=\frac{\pi}{6}e^{-\frac{1}{2}iy}e^{-\frac{2}{3}|y|}\Rightarrow I_1+I_2=\frac{\pi}{6}e^{-\frac{3}{2}i}\cdot e^{-2}\\ &I_1=\text{Re}(I_1+I_2)=\frac{\pi}{6e^2}\cos\frac{3}{2} \end{aligned} $$ However, the correct answer is this: $$I_1=\frac{\pi}{6e^2}\cos 1$$
I've tried different approaches, and yet I don't realize what is wrong with my solution. Any help will be appreciated!
Mistake when changing variables? $$F\left[\frac{1}{\left(\frac{3x}{2}-\left(-\frac{1}{2}\right)\right)^2 +1}\right]=e^{-\frac{1}{3}iy}\cdot\frac{2}{3}\pi e^{-\frac{2}{3}|y|}$$