Evaluation of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\dotsb$

How can we calculate sum of following infinite series

$\displaystyle \bullet\; 1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$

$\displaystyle \bullet\; 1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\cdots$

$\textbf{My Try:}$ Let $$S = \int_0^1 (1+x^2-x^4-x^6+x^8+x^{10}+\cdots) \, dx$$

So we get $$S=\int_0^1 \left(1-x^4+x^8-\cdots\right)dx+x^2\int_0^1 (1-x^4+x^8-\cdots)\,dx$$

So we get $$S=\int_0^1 \frac{1+x^2}{1+x^4} \, dx= \frac{\pi}{2\sqrt{2}}$$ after that we can solve it

Again for second One, Let $$S=\int_0^1 (1-x^6+x^8-x^{14}+x^{16}+\cdots)$$

So we get $$S=\int_0^1 (1+x^8+x^{16}+\cdots) \, dx-\int_0^1 (x^6+x^{14}+ \cdots)\,dx$$

So we get $$S=\int_{0}^{1}\frac{1-x^6}{1-x^8}dx = \int_{0}^{1}\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$

Now how can i solve after that, Help me

Thanks

Start with partial fractions

$$\frac{x^4+x^2+1}{(x^2+1)(x^4+1)} = \frac{A}{x^2+1}+ \frac{B x^2+C}{x^4+1}$$

Thus,

$$A+B=1$$ $$B+C=1$$ $$A+C=1$$

or $A=B=C=1/2$. Also note that

$$x^4+1 = (x^2+\sqrt{2} x+1)(x^2-\sqrt{2} x+1) $$

so that

$$\frac{x^2+1}{x^4+1} = \frac{P}{x^2-\sqrt{2} x+1} + \frac{Q}{x^2+\sqrt{2} x+1}$$

where $P=Q=1/2$. Thus,

$$\frac{x^4+x^2+1}{(x^2+1)(x^4+1)} = \frac14 \left [2 \frac1{x^2+1} + \frac1{(x-\frac1{\sqrt{2}})^2+\frac12} + \frac1{(x+\frac1{\sqrt{2}})^2+\frac12} \right ]$$

And the integral is

$$\frac12 \frac{\pi}{4}+ \frac14 \sqrt{2} \left [\arctan{(\sqrt{2}-1)}-\arctan{(-1)} \right ] + \frac14 \sqrt{2} \left [\arctan{(\sqrt{2}+1)}-\arctan{(1)} \right ]= \frac{\pi}{8} (\sqrt{2}+1) $$

For the second, you're doing fine. You need to use partial fractions for the integral you've gotten. $x^2 + 1$ is irreducible, but the $x^4 + 1$ may be giving you troubles.

Hint: $$x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1) $$