Calculate ∫01arctanxx dx\int_0^1 \frac{\arctan x}{x} \, dx∫01xarctanxdx expressed as a series.
∑n=0∞(−1)n(2n+1)2\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}∑n=0∞(2n+1)2(−1)n
∑n=0∞(−1)n2n+1\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}∑n=0∞2n+1(−1)n
∑n=1∞(−1)nn2\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}∑n=1∞n2(−1)n
∑n=0∞1(2n+1)2\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}∑n=0∞(2n+1)21