Given the function $f(x,t)$, solve the following PDE
$$\partial_{tt}f+ 2\partial_{t}f- \partial_{xx}f+f=0$$
BC: $$f(x=\pm \infty, t)=0$$ IC: $$f(x,t=0)=g(x), \quad f_t(x,t=0)=0$$
I tried solving it using FT in $x$
$$\partial_{tt}F+ 2\partial_{t}F-(1+w^2)F=0$$
then solving the characteristic equation
$$F=C_{1} \exp(-1+iw)+ C_{2} \exp(-1-iw)$$
Using the IC $$F=G(w) \left(\frac{(iw+1) \exp(-1+iw) + (iw-1) \exp(-1-iw)}{2iw} \right)$$
after this I'm stuck because I couldn't inverse FT and find $f$ I appreciate any help I could get.
Let's take a shorter route. Defining $v(x,t)=e^tf(x,t)$, it is straightfoward to show that $v$ satisfies the wave equation $\partial_{tt}v-\partial_{xx}v=0$, which has the general solution $v(x,t)=u(x+t)+w(x-t)$, where $u$ and $v$ are arbitrary differentiable functions. Hence, the general solution to the original PDE is $$ f(x,t)=e^{-t}[u(x+t)+w(x-t)]. \tag{1} $$ The initial conditions yield the system of equations $$ \begin{cases} u(x)+w(x)=g(x), \\ u'(x)-w'(x)-u(x)-w(x)=0, \tag{2} \end{cases} $$ which has as solution \begin{align} u(x)&=\frac{1}{2}g(x)+\frac{1}{2}\int_{x_0}^{x}g(\xi)\,d\xi, \tag{3a} \\ w(x)&=\frac{1}{2}g(x)-\frac{1}{2}\int_{x_0}^{x}g(\xi)\,d\xi. \tag{3b} \end{align} Plugging $(3)$ into $(1)$, we finally obtain $$ f(x,t)=\frac{1}{2}e^{-t}\left(g(x+t)+g(x-t)+\int_{x-t}^{x+t}g(\xi)\,d\xi \right). \tag{4} $$
ADDENDUM
Let me complement my answer by showing how to arrive at Eq. $(4)$ from your partial solution. First, let's add a $t$ to the exponentials (which you forgot when you solved the equation $\partial_{tt}F+ 2\partial_{t}F-(1+w^2)F=0$), and rearrange terms: \begin{align} F(w,t)&=G(w) \frac{(iw+1) e^{(-1+iw)t} + (iw-1) e^{(-1-iw)t}}{2iw} \\ &=\frac{1}{2}e^{-t}G(w)\left[e^{iwt}+e^{-iwt}+\frac{1}{iw} \left(e^{iwt}-e^{-iwt}\right)\right]. \tag{A.1} \end{align} From $(A.1)$ we obtain $f(x,t)$ as $$ f(x,t)=\int_{-\infty}^{\infty}F(w,t)e^{iwx}dw =\frac{1}{2}e^{-t}\left(I_1+I_2+I_3\right), \tag{A.2} $$ where each $I_k$ corresponds to the integral of a term in $(A.1)$: \begin{align} I_1&=\int_{-\infty}^{\infty}G(w)e^{iw(x+t)}dw=g(x+t), \tag{A.3a} \\ I_2&=\int_{-\infty}^{\infty}G(w)e^{iw(x-t)}dw=g(x-t), \tag{A.3b} \\ I_3&=\int_{-\infty}^{\infty}G(w)\frac{1}{iw}\left(e^{iw(x+t)}-e^{iw(x-t)}\right)dw \\ &=\int_{-\infty}^{\infty}G(w)\int_{x-t}^{x+t}e^{iw\xi}\,d\xi\, dw \\ &=\int_{x-t}^{x+t}\int_{-\infty}^{\infty}G(w)e^{iw\xi}\,dw\,d\xi \\ &=\int_{x-t}^{x+t}g(\xi)\,d\xi. \tag{A.3c} \end{align} Combining $(A.2)$ and $(A.3)$ we obtain the solution $(4)$.