Integrating $\int_{-\infty}^{\infty} \mathcal{N}(\mu,\sigma^2) \log(\cosh(x)) dx$ by parts

Question

I need to verify this integration (by parts): $I=\int_{-\infty}^{\infty} \mathcal{N}(\mu,\sigma^2)*\log(\cosh(x)) dx\\$

$u=\log(\cosh(x))$ ,then $du=\tanh(x) dx$

$\\dv=\mathcal{N}(\mu,\sigma^2) dx$ ,then $v=1$

using the formula of By parts:$I=uv-\int_{-\infty}^{\infty}v du$

$I=\log(\cosh(x))-\int_{-\infty}^{\infty}\tanh(x) dx$

$I=\log(\cosh(x))-\log(\cosh(x))=0$