Evaluate ∫01arctan(x)xdx\int_0^1 \frac{\arctan(x)}{x} dx∫01xarctan(x)dx.
∑n=0∞(−1)n(2n+1)2\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}∑n=0∞(2n+1)2(−1)n
π4\frac{\pi}{4}4π
ln(2)\ln(2)ln(2)
12\frac{1}{2}21