You can use the formula for finding residues of a pole $z_0$ of order $1$:
$$\text{Res}(f, z_0)=\lim \limits_{z\to z_0}\left[(z-z_0)f(z)\right].$$
The singularities here are the points $(2k+1)\pi i$, where $k$ ranges over the integers.
To use the above formula you need to first prove that these points are poles of order $1$, that is, you need to prove that these points are not removable discontinuities and you need to prove that $\lim \limits_{z\to (2k+1)\pi i}\left(\dfrac{z-(2k+1)\pi i}{e^z+1}\right)\in \mathbb C$. This limit actually is the same as in the above formula, so if you find it, you immediately find your residue.
One has
$$\begin{align} \lim \limits_{z\to (2k+1)\pi i}\left(\dfrac{z-(2k+1)\pi i}{e^z+1}\right)&=\lim \limits_{w\to 0}\left(\dfrac{w}{e^{w+(2k+1)\pi i}+1}\right)\\
&=\lim \limits_{w\to 0}\left(\dfrac{w}{e^{w+\pi i}+1}\right)\\
&=\lim \limits_{w\to 0}\left(\dfrac{1}{e^{w+\pi i}}\right)\\
&=-1.
\end{align}$$
Hence $\forall k\in \mathbb Z\left(\text{Res}(f,(2k+1)\pi i)=-1\right)$ which agrees with Wolfram Alpha.
In the above I used Cauchy's rule. If you want to avoid without having to resort to Laurent series, you can first prove that $\lim \limits_{w\to 0}\left(\dfrac{e^{w+\pi i}+1}w\right)=1$ and this you can do with the Taylor Series.