Integral $\int_{-\infty}^{\infty} \frac{1-x}{1-x^5}\, d x $

Question

I want to compute the integral,
$$ \mathcal{I} = \int_{-\infty}^{\infty} \frac{1-x}{1-x^5}\, d x $$ And my attempt was like this,
$$ \begin{align*} \mathcal{I} &= \int_{-\infty}^{\infty} \frac{1-x}{1-x^5}\, d x \\ &= \underbrace{\int_{0}^{\infty} \frac{1+x}{1+x^5}\, d x}_{\mathbf{I_1}} + \underbrace{\int_{0}^{\infty} \frac{1-x}{1-x^5}\, d x}_{\mathbf{I_2}} \\ \end{align*} $$
I solved $\mathbf{I}_1 $ using Beta Function
$$ \begin{align*} \mathbf{I}_1 &= \int_{0}^{\infty} \frac{1+x}{1+x^5}\, d x \\ &= \frac{1}{5} \int_0^\infty \frac{1 + t^{1/5}}{(1 + t)} \, t^{-4/5} \, d t && \text{($ x^5 \rightarrow t $)} \\ &= \frac{1}{5} \left( \int_0^\infty \frac{t^{-3/5}}{(1 + t)} \, d t + \int_0^\infty \frac{ t^{-4/5}}{(1 + t)} \, d t \right) \end{align*} $$ We know that ,
$$ \beta{(m, n)} = \int_0^\infty \frac{ x^{n-1}}{(1 + x)^{m+n}} \, d x $$ $$ \beta{(m, n)} = \frac{\Gamma{(m)}\,\Gamma{(n)}}{\Gamma{(m+n)}} $$
And $$ \Gamma{(x)}\,\Gamma{(1-x)} = \frac{\pi}{\sin(\pi x)} $$ So,
$$ \begin{align*} \mathbf{I}_1 &= \frac{1}{5} \left(\beta{(1/5, 4/5)} + \beta{(2/5, 3/5)} \right) \\ &= \frac{\pi}{5} \left( \frac{1}{\sin \left(\frac{\pi}{5} \right)} + \frac{1}{\sin \left(\frac{2 \pi}{5} \right)} \right) \end{align*} $$,
But I have no idea how to solve
$$ \mathbf{I}_2 = \int_{0}^{\infty} \frac{1-x}{1-x^5}\, d x $$
I tried some substitution like $ x^5 = \sin^2(\theta) $ But didn't work.
Would you please help me ?
Thank you !