Thanks for the hints by @whuber. Let me answer this question myself.

The quadratic term can be written as $$ \sum_{i=1}^{N}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta})^{\top} \boldsymbol{V}^{-1}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta}) =\sum_{i=1}^{N}\boldsymbol{y}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i} -2\boldsymbol{\beta}^{\top}\left(\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top} \boldsymbol{V}^{-1}\boldsymbol{y}_{i}\right) +\boldsymbol{\beta}^{\top}\left(\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top} \boldsymbol{V}^{-1}\boldsymbol{X}_{i}\right)\boldsymbol{\beta} $$

Let $\boldsymbol{A}=\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{X}_{i}$, $\boldsymbol{b}=\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i}$, $c=\sum_{i=1}^{N}\boldsymbol{y}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i}$. Here, $\boldsymbol{A}$ is a $K \times K$ symmetric positive matrix, $\boldsymbol{b}$ is a $K\times 1$ vector, and $c$ is a scalar.

Now the integral term is

$$ \exp\left(-\frac{1}{2}c\right)\int_{\mathbb{R}^{K}} \exp\left(-\frac{1}{2}\boldsymbol{\beta}^{\top}\boldsymbol{A}\boldsymbol{\beta}+ \boldsymbol{b}^{\top}\boldsymbol{\beta}\right)d\boldsymbol{\beta} $$

Applying the Gaussian integral formula (n-dimensional with linear term) from Wikipedia, the integral term is

$$ \exp\left(-\frac{1}{2}c\right) \exp\left(\frac{1}{2}\boldsymbol{b}^{\top}\boldsymbol{A}^{-1}\boldsymbol{b}\right) \sqrt{\frac{(2\pi)^{K}}{\det(\boldsymbol{A})}} $$

Thanks for the hints by @whuber. Let me answer this question myself.

The quadratic term can be written as $$ \sum_{i=1}^{N}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta})^{\top} \boldsymbol{V}^{-1}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta}) =\sum_{i=1}^{N}\boldsymbol{y}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i} -2\boldsymbol{\beta}^{\top}\left(\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top} \boldsymbol{V}^{-1}\boldsymbol{y}_{i}\right) +\boldsymbol{\beta}^{\top}\left(\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top} \boldsymbol{V}^{-1}\boldsymbol{X}_{i}\right)\boldsymbol{\beta} $$

Let $\boldsymbol{A}=\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{X}_{i}$, $\boldsymbol{b}=\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i}$, $c=\sum_{i=1}^{N}\boldsymbol{y}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i}$. Here, $\boldsymbol{A}$ is a $K \times K$ symmetric positive definite matrix, $\boldsymbol{b}$ is a $K\times 1$ vector, and $c$ is a scalar.

Now the integral term is

$$ \exp\left(-\frac{1}{2}c\right)\int_{\mathbb{R}^{K}} \exp\left(-\frac{1}{2}\boldsymbol{\beta}^{\top}\boldsymbol{A}\boldsymbol{\beta}+ \boldsymbol{b}^{\top}\boldsymbol{\beta}\right)d\boldsymbol{\beta} $$

Applying the Gaussian integral formula (n-dimensional with linear term) from Wikipedia, the integral term is

$$ \exp\left(-\frac{1}{2}c\right) \exp\left(\frac{1}{2}\boldsymbol{b}^{\top}\boldsymbol{A}^{-1}\boldsymbol{b}\right) \sqrt{\frac{(2\pi)^{K}}{\det(\boldsymbol{A})}} $$

Thanks for the hints by @whuber. Let me answer this question myself.

The quadratic term can be written as $$ \sum_{i=1}^{N}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta})^{\top} \boldsymbol{V}^{-1}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta}) =\sum_{i=1}^{N}\boldsymbol{y}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i} -2\boldsymbol{\beta}^{\top}\left(\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top} \boldsymbol{V}^{-1}\boldsymbol{y}_{i}\right) +\boldsymbol{\beta}^{\top}\left(\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top} \boldsymbol{V}^{-1}\boldsymbol{X}_{i}\right)\boldsymbol{\beta} $$

Let $boldsymbol{A}=\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{X}_{i}$, $boldsymbol{b}=\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i}$, $c=\sum_{i=1}^{N}\boldsymbol{y}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i}$. Here, $boldsymbol{A}$ is a $K \times K$ symmetric positive matrix, $boldsymbol{b}$ is a $K\times 1$ vector, and $c$ is a scalar.

Now the integral term is

$$ \exp\left(-\frac{1}{2}c\right)\int_{\mathbb{R}^{K}} \exp\left(-\frac{1}{2}\boldsymbol{\beta}^{\top}\boldsymbol{A}\boldsymbol{\beta}+ \boldsymbol{b}^{\top}\boldsymbol{\beta}\right)d\boldsymbol{\beta} $$

Applying the Gaussian integral formula (n-dimensional with linear term) from Wikipedia, the integral term is

$$ \exp\left(-\frac{1}{2}c\right) \exp\left(\frac{1}{2}\boldsymbol{b}^{\top}\boldsymbol{A}^{-1}\boldsymbol{b}\right) \sqrt{\frac{(2\pi)^{K}}{\det(\boldsymbol{A})}} $$

Thanks for the hints by @whuber. Let me answer this question myself.

The quadratic term can be written as $$ \sum_{i=1}^{N}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta})^{\top} \boldsymbol{V}^{-1}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta}) =\sum_{i=1}^{N}\boldsymbol{y}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i} -2\boldsymbol{\beta}^{\top}\left(\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top} \boldsymbol{V}^{-1}\boldsymbol{y}_{i}\right) +\boldsymbol{\beta}^{\top}\left(\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top} \boldsymbol{V}^{-1}\boldsymbol{X}_{i}\right)\boldsymbol{\beta} $$

Let $\boldsymbol{A}=\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{X}_{i}$, $\boldsymbol{b}=\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i}$, $c=\sum_{i=1}^{N}\boldsymbol{y}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i}$. Here, $\boldsymbol{A}$ is a $K \times K$ symmetric positive matrix, $\boldsymbol{b}$ is a $K\times 1$ vector, and $c$ is a scalar.

Now the integral term is

$$ \exp\left(-\frac{1}{2}c\right)\int_{\mathbb{R}^{K}} \exp\left(-\frac{1}{2}\boldsymbol{\beta}^{\top}\boldsymbol{A}\boldsymbol{\beta}+ \boldsymbol{b}^{\top}\boldsymbol{\beta}\right)d\boldsymbol{\beta} $$

Applying the Gaussian integral formula (n-dimensional with linear term) from Wikipedia, the integral term is

$$ \exp\left(-\frac{1}{2}c\right) \exp\left(\frac{1}{2}\boldsymbol{b}^{\top}\boldsymbol{A}^{-1}\boldsymbol{b}\right) \sqrt{\frac{(2\pi)^{K}}{\det(\boldsymbol{A})}} $$