Series $\sum_{n=1}^{\infty}(-1)^n\frac{n}{(2n-1)^2}$

Define $$f(x)=\sum_{n=1}^\infty \frac{(-1)^n x^{2n}}{2n-1}=-x\arctan(x)$$ Then differentiate this to obtain $$f'(x)=\sum_{n=1}^\infty \frac{(-1)^n 2n x^{2n-1}}{2n-1}=-\arctan(x)-\frac{x}{x^2+1}$$ Divide both sides by $x$ to obtain $$\frac{f'(x)}{x}=\sum_{n=1}^\infty \frac{(-1)^n 2n x^{2n-2}}{2n-1}=-\frac{\arctan(x)}{x}-\frac{1}{x^2+1}$$ Then integrate both sides from $x=0$ to $1$, giving $$\int_0^1\frac{f'(x)}{x}dx =\sum_{n=1}^\infty \frac{(-1)^n 2n}{(2n-1)^2}=-\int_0^1\frac{\arctan(x)}{x}dx-\int_0^1\frac{dx}{x^2+1}$$ The first integral on the RHS is equal to $G$, the Catalan Constant, which can be shown by expanding the arctangent in the integrand with its Taylor Series. The second integrand has an elementary antiderivative. Thus, we have $$\int_0^1\frac{f'(x)}{x}dx =\sum_{n=1}^\infty \frac{(-1)^n 2n}{(2n-1)^2}=-G-\frac{\pi}{4}$$ or $$\sum_{n=1}^\infty \frac{(-1)^n n}{(2n-1)^2}=-\frac{G}{2}-\frac{\pi}{8}$$

The sum is easily evaluated by recognizing that

$$\frac{n}{(2 n-1)^2} = \frac12 \frac1{2 n-1} + \frac12 \frac1{(2 n-1)^2} $$