Find the value of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$

HINT: Notice that $$\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=\frac{1}{4}\frac{1}{x^2+x\sqrt{2}+1}+\frac{1}{4}\frac{1}{x^2-x\sqrt{2}+1}+\frac{1}{2}\frac{1}{x^2+1}$$ Now we just have to evaluate the integrals $$I_1=\frac{1}{4}\int_0^1 \frac{dx}{x^2+x\sqrt{2}+1}$$ $$I_2=\frac{1}{4}\int_0^1 \frac{dx}{x^2-x\sqrt{2}+1}$$ $$I_3=\frac{1}{2}\int_0^1 \frac{dx}{x^2+1}$$ and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?

Another approach, which could be useful to check yours through integrals, is same as per the similar post, using the Digamma function $\psi (z)$. $$ \eqalign{ & 1 - {1 \over {8 - 1}} + {1 \over {8 + 1}} - {1 \over {2 \cdot 8 - 1}} + {1 \over {2 \cdot 8 + 1}} + \cdots = \cr & = 1 - \sum\limits_{1\, \le \,n} {{1 \over {8n - 1}}} + \sum\limits_{1\, \le \,n} {{1 \over {8n + 1}}} = \cr & = 1 - {1 \over 8}\left( {\sum\limits_{1\, \le \,n} {{1 \over {n - 1/8}}} - \sum\limits_{1\, \le \,n} {{1 \over {n + 1/8}}} } \right) = \cr & = 1 - {1 \over 8}\left( {\sum\nolimits_{\;n = 7/8}^{\;\infty } {{1 \over n}} - \sum\nolimits_{\;n = 9/8}^{\;\infty } {{1 \over n}} } \right) = \cr & = 1 - {1 \over 8}\left( {\sum\nolimits_{\;n = 7/8}^{\;9/8} {{1 \over n}} } \right) = \cr & = 1 - {1 \over 8}\left( {\psi (9/8) - \psi (7/8)} \right) = \cr & = 1 - {1 \over 8}\left( {\psi \left( {1 + 1/8} \right) - \psi \left( {1 - 1/8} \right)} \right) = \cr & = 1 - {1 \over 8}\left( {{1 \over {1/8}} + \psi \left( {1/8} \right) - \psi \left( {1 - 1/8} \right)} \right) = \cr & = {1 \over 8}\left( {\psi \left( {1 - 1/8} \right) - \psi \left( {1/8} \right)} \right) = \cr & = {1 \over 8}\left( {\pi \cot \left( {{\pi \over 8}} \right)} \right) = \cr & = {{\pi \left( {1 + \sqrt 2 } \right)} \over 8} \cr} $$ where:

$$ \Delta _{\,z} \psi \left( z \right) = \psi \left( {z + 1} \right) - \psi \left( z \right) = {1 \over z} $$ is the functional equation for the Digamma;
which implies that Digamma is the Antidelta of $1/z$ $$ \eqalign{ & \psi \left( z \right) = \Delta _{\,z} ^{ - 1} \left( {{1 \over z}} \right) = \sum\nolimits_z {{1 \over z}} \quad \Rightarrow \cr & \Rightarrow \quad \sum\nolimits_{\;z = a}^{\;b} {{1 \over z}} = \psi \left( b \right) - \psi \left( a \right) \cr} $$ and we used the Reflection formula for Digamma $$ \psi \left( {1 - z} \right) = \psi \left( z \right) + \pi \cot \left( {\pi z} \right) $$

Now, the above, suggests a way to solve the integral.

Let's replace $x^8$ with $y$ $$ \int_{x = 0}^{\,1} {{{1 - x^{\,6} } \over {1 - x^{\,8} }}dx} \quad \mathop = \limits^{x^{\,8} = y} \quad {1 \over 8}\int_{y = 0}^{\,1} {{{1 - y^{\,3/4} } \over {\left( {1 - y} \right)y^{\,7/8} }}dy} $$ then consider that we have $$ \eqalign{ & {{1 - y^{\,3/4} } \over {\left( {1 - y} \right)y^{\,7/8} }} = \cr & = \left( {\left( {1 - y} \right)^{ - 1} y^{\, - 7/8} - \left( {1 - y} \right)^{ - 1} y^{\, - 1/8} } \right) \cr & = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {\left( {1 - y} \right)^{\varepsilon - 1} y^{\,1/8 - 1} - \left( {1 - y} \right)^{\varepsilon - 1} y^{\,\,7/8 - 1} } \right) \cr} $$ so we are ready to use the integral and Gamma representation for the Beta function $$ \eqalign{ & 8 \;\int_{x = 0}^{\,1} {{{1 - x^{\,6} } \over {1 - x^{\,8} }}dx} = \int_{y = 0}^{\,1} {{{1 - y^{\,3/4} } \over {\left( {1 - y} \right)y^{\,7/8} }}dy} = \cr & = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {\int_{y = 0}^{\,1} {\left( {1 - y} \right)^{\varepsilon - 1} y^{\,1/8 - 1} dy} - \int_{y = 0}^{\,1} {\left( {1 - y} \right)^{\varepsilon - 1} y^{\,\,7/8 - 1} dy} } \right) = \cr & = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {{\rm B}\left( {1/8,\varepsilon } \right) - {\rm B}\left( {7/8,\varepsilon } \right)} \right) = \cr & = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {{{\Gamma \left( {1/8} \right)\Gamma \left( \varepsilon \right)} \over {\Gamma \left( {1/8 + \varepsilon } \right)}} - {{\Gamma \left( {7/8} \right)\Gamma \left( \varepsilon \right)} \over {\Gamma \left( {7/8 + \varepsilon } \right)}}} \right) = \cr & = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {{{\Gamma \left( {1/8} \right)\Gamma \left( \varepsilon \right)} \over {\psi \left( {1/8} \right)\varepsilon \;\Gamma \left( {1/8} \right) + \Gamma \left( {1/8} \right)}} - {{\Gamma \left( {7/8} \right)\Gamma \left( \varepsilon \right)} \over {\psi \left( {7/8} \right)\varepsilon \;\Gamma \left( {7/8} \right) + \Gamma \left( {7/8} \right)}}} \right) = \cr & = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \left( {{{\Gamma \left( \varepsilon \right)} \over {\psi \left( {1/8} \right)\varepsilon \; + 1}} - {{\Gamma \left( \varepsilon \right)} \over {\psi \left( {7/8} \right)\varepsilon \; + 1}}} \right) = \cr & = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \Gamma \left( \varepsilon \right)\left( {\left( {1 - \psi \left( {1/8} \right)\varepsilon \;} \right) - \left( {1 - \psi \left( {7/8} \right)\varepsilon \;} \right)} \right) = \cr & = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \varepsilon \,\Gamma \left( \varepsilon \right)\left( {\psi \left( {7/8} \right) - \psi \left( {1/8} \right)} \right) = \cr & = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \,\Gamma \left( {1 + \varepsilon } \right)\left( {\psi \left( {7/8} \right) - \psi \left( {1/8} \right)} \right) = \cr & = \psi \left( {7/8} \right) - \psi \left( {1/8} \right) \cr} $$ which is the same result as above.

Let me add a more straight derivation of the above.

We rewrite the integrand as $$ \eqalign{ & {{1 - y^{\,3/4} } \over {\left( {1 - y} \right)y^{\,7/8} }} = {{y^{\, - 7/8} - y^{\, - 1/8} } \over {\left( {1 - y} \right)}} = \cr & = {{1 - y^{\, - 1/8} - \left( {1 - y^{\, - 7/8} } \right)} \over {\left( {1 - y} \right)}} \cr} $$ and compare with the integral representation of Digamma $$ \psi (s + 1) = - \gamma + \int_0^1 {{{1 - x^{\,s} } \over {1 - x}}dx} $$

The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+\sqrt 2 x+1)(x^2-\sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $\frac{1}{4}\left(2\tan^{-1}x+\sqrt{2}\left(\tan^{-1}\left(\sqrt{2}x+1\right)-\tan^{-1}\left(1-\sqrt{2}x\right)\right)\right)$