The Nelson-Siegel (1987) model states that the instantaneous forward rate at maturity, $r(m)$ is given as: $$\beta_0 + \beta_1 e^{(-m/\tau)} + \beta_2\left(\frac{m}{\tau}e^{(-m/\tau)}\right) \tag{1} $$ The yield to maturity, $R(m)$ is given by: $$R(m) = \frac{1}{m}\int_0^mr(x)dx \tag{2} $$
On solving $(2)$ I get the following as $R(m)$: $$ R(m) = \beta_0 + \frac{(\beta_1+\beta_2)(1- e^{(-m/\tau)})}{\frac{m}{\tau}} - \beta_2\frac{e^{(-m/\tau)}}{\tau} \tag{3} $$
However the correct equation for $R(m)$ is given in the paper as: $$ R(m) = \beta_0 + \frac{(\beta_1+\beta_2)(1- e^{(-m/\tau)})}{\frac{m}{\tau}} - \beta_2e^{(-m/\tau)} \tag{4} $$
I feel like I did the integration wrong but no matter what, I simply cannot get the $\tau$ to disappear in the last term of $(3)$. Any idea on what I am doing wrong?
The result in the paper is correct as written. Since you have not provided the details of your computation, I cannot tell you why you have an extra $\tau$ in the denominator of the $\beta_2$ term. If you would like to have such an explanation, you will need to furnish the intermediate calculations that resulted in $(3)$.
For the sake of simplicity, let us define
$$f(x) = \beta_0 + (\beta_1 + \beta_2 x)e^{-x}.$$ Then $f(m/\tau) = r(m)$. We also observe that $$\frac{d}{dx}[xe^{-x}] = e^{-x} + x(-e^{-x}) = (1-x)e^{-x}.$$ This suggests writing $$f(x) = \beta_0 + (\beta_1 + \beta_2)e^{-x} - \beta_2(1 - x)e^{-x},$$ so that $$\int f(x) \, dx = \beta_0 x - (\beta_1 + \beta_2)e^{-x} - \beta_2 x e^{-x} + C.$$
Consequently, with the substitution $x = \tau u$, $$\begin{align} R(m) &= \frac{1}{m} \int_{x=0}^m r(x) \, dx \\ &= \frac{\tau}{m} \int_{u=0}^{m/\tau} r(\tau u) \, du \\ &= \frac{\tau}{m} \int_{u=0}^{m/\tau} f(u) \, du \\ &= \frac{\tau}{m} \left( \beta_0 \frac{m}{\tau} - (\beta_1 + \beta_2) (e^{-m/\tau} - 1) - \beta_2 \frac{m}{\tau} e^{-m/\tau}\right) \\ &= \beta_0 + \frac{(\beta_1 + \beta_2)(1-e^{-m/\tau})}{m/\tau} - \beta_2 e^{-m/\tau} \end{align}$$ as claimed.