How I can calculate $\sum_{n = 0}^{\infty}\frac{(-1)^n(2n + 1)}{n^2 + 1}$

Question

I turned a series into the integral $$ \int_{0}^{1}\frac{\cos(\log x)}{x^2 + 1}dx $$ using partial fractions and other techniques, but I don't know what to do next. I did a search on the site but found nothing similar. Thus, $$ \sum_{n = 0}^{\infty}\frac{(-1)^n(2n + 1)}{n^2 + 1} = \int_{0}^{1}\frac{\cos(\log x)}{x^2 + 1}dx = ? $$