What is the value of $\frac11+\frac13-\frac15-\frac17+\frac19+\frac1{11}-\dots$?

Let $$\begin{align} S_1&=1-\frac15+\frac19-{1\over13}+\dots\\ S_2&=\frac13-\frac17+{1\over11}-{1\over15}+\dots \end{align}$$ so that the sum we seek is $S_1+S_2.$ T0 compute $S_1,$ consider $$f(x) = 1-{x^5\over5}+{x^9\over9}-{x^{13}\over13}+\dots$$ so that $$f'(x)=-x^4+x^8-x^{12}+\dots={-x^4\over1+x^4},\ |x|<1$$ and $$f(x)=\int_0^x{-t^4\over1+t^4}\mathrm{dt}+f(0),\ |x|<1$$ By Abel's limit theorem, $$S_1=\lim_{x\to1-}\int_0^x{-t^4\over1+t^4}\mathrm{dt}+f(0)=1-\int_0^1{t^4\over1+t^4}\mathrm{dt}$$ and we can do a similar calculation for $S_2$ to get $$S_2=\int_0^1{t^2\over1+t^4}\mathrm{dt}$$

The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.

Hint:

$$\frac{1}{1}+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\dots=\sum_{k=1}^{\infty }\frac{1}{8k-7}+\frac{1}{8k-5}-\frac{1}{8k-3}-\frac{1}{8k-1}$$

$$=\sum_{k=1}^{\infty }(\frac{1}{8k-7}-\frac{1}{8k-1})+(\frac{1}{8k-5}-\frac{1}{8k-3})$$

$$=\sum_{k=1}^{\infty }(1+\frac{1}{8k+1}-\frac{1}{8k-1})+(\frac{1}{3}+\frac{1}{8k+3}-\frac{1}{8k-3})$$

$$=\frac{4}{3}+\sum_{k=1}^{\infty }(\frac{-2}{64k^2-1}+\frac{-6}{64k^2-9})$$ $$=\frac{4}{3}-\frac{1}{32}\sum_{k=1}^{\infty }\frac{1}{k^2-\frac{1}{64}}-\frac{3}{32}\sum_{k=1}^{\infty }\frac{1}{k^2-\frac{9}{64}}$$

then use $$\frac{1-\pi x \cot(\pi x)}{2x^2}=\sum_{k=1}^{\infty }\frac{1}{k^2-x^2}$$

$$ \begin{align} &\sum_{k=0}^\infty\left(\frac1{8k+1}+\frac1{8k+3}-\frac1{8k+5}-\frac1{8k+7}\right)\\ &=\sum_{k=0}^\infty\left(\frac1{8k+1}-\frac1{8k+7}\right)+\sum_{k=0}^\infty\left(\frac1{8k+3}-\frac1{8k+5}\right)\tag1\\ &=\sum_{k\in\mathbb{Z}}\frac1{8k+1}+\sum_{k\in\mathbb{Z}}\frac1{8k+3}\tag2\\ &=\frac18\sum_{k\in\mathbb{Z}}\frac1{k+\frac18}+\frac18\sum_{k\in\mathbb{Z}}\frac1{k+\frac38}\tag3\\ &=\frac\pi8\left[\cot\left(\frac\pi8\right)+\cot\left(\frac{3\pi}8\right)\right]\tag4\\[6pt] &=\frac{\pi\sqrt2}4\tag5 \end{align} $$ Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $\mathbb{Z}$
$(3)$: factor $\frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $\cot\left(\frac\pi8\right)=1+\sqrt2$ and $\cot\left(\frac{3\pi}8\right)=-1+\sqrt2$