# Double Integral of Bump Function

I can't figure out how to show that the following equality holds.

$$\int_{-1}^1 \int_{-1}^x \exp\left(\frac{1}{y^2 - 1}\right) dy dx = \int_{-1}^1 \exp\left(\frac{1}{y^2 - 1}\right) dy$$

Any help or hint is appreciated. I tried to use my standard integration toolbox, but I didn't succeed.

\begin{align} \int_{-1}^1 \int_{-1}^x f(y)\,dy\,dx &= \int_{-1}^1\int_{y}^1 f(y)\,dx\,dy\\ &= \int_{-1}^1 (1-y)\,f(y)\,dy\\ &= \int_{-1}^1 f(y)\, dy - \int_{-1}^1 yf(y)\,dy. \end{align}
In our case, the function $f(y) = \exp \left(\frac{1}{y^2-1}\right)$ is even, hence $yf(y)$ is odd, and the second integral vanishes, leaving only the first.