Fourier transform relationship
Consider the signal $$s[n]=\frac{a^{|n|}}{1-a^2},\, |a|&lt;1$$

Assuming $$x[n]=a^n u[n]\stackrel{\text{DTFT}}\longleftrightarrow\frac{1}{1-a e^{-j\omega}},$$
 one can show that 

$$s[n]=x[n]*x[-n]$$

The proof is simple. For example, for $n&gt;0$ we have
$$\sum_{k=n}^{\infty}x[k]x[k-n]=\sum_{k=n}^{\infty}a^ka^{k-n}=\sum_{k=n}^{\infty}a^{2k-n}=a^n(1+a^2+a^4+\cdots)=\frac{a^n}{1-a^2}$$

Hence, the DTFT of the signal $s[n]$ is given by
$$S(e^{j\omega})=X(e^{j\omega})X(e^{-j\omega})=\frac{1}{1-a e^{-j\omega}}\frac{1}{1-a e^{j\omega}}=\frac{1}{1-2a\cos(\omega)+a^2}$$
In summary,
$$\boxed{\frac{a^{|n|}}{1-a^2}\stackrel{\text{DTFT}}\longleftrightarrow\frac{1}{1-2a\cos(\omega)+a^2},\, |a|&lt;1}$$

Now consider 

$$G(f)=\frac{1}{1+1.4\cos(2\pi f)+0.49}=\frac{1}{(1+0.7e^{-j2\pi f})(1+0.7e^{j2\pi f})}$$

whose IDTFT can be calculated using the above formula for $a=-0.7$ as

$$g[n]=\frac{(-0.7)^{|n|}}{1-(-0.7)^2}$$

Since 
$$\frac{1+0.8(e^{-j 2\pi f}+e^{j 2\pi f})+0.64}{1+1.4\cos(2\pi f)+0.49}=G(f)+0.8e^{-j 2\pi f}G(f)+0.8e^{j 2\pi f}G(f)+0.64G(f)$$
The IDTFT is
$$g[n]+0.8g[n-1]+0.8g[n+1]+0.64g[n]$$
But obviously, the variable $\tau$ is used instead of $n$ in the give equation.