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Find the value of this :$$1-frac17+frac19-frac115+frac117-frac123+frac125....$$

Try: We can write the above series as

$$S = int^1_0bigg[1-x^6+x^8-x^14+x^16-x^22+cdotsbigg]dx$$

$$S = int^1_0(1-x^6)bigg[1+x^8+x^16+cdots cdots bigg]dx$$

So $$S = int^1_0frac1-x^61-x^8dx = int^1_0fracx^4+x^2+1(x^2+1)(x^4+1)dx$$

Now i am struck in that integration.

Did not understand how to solve it

could some help me to solve it. Thanks in advance

Another approach, which could be useful to check yours through integrals,
is same as per the similar post, using the Digamma function $psi (z)$.
$$
eqalign
& 1 - 1 over 8 - 1 + 1 over 8 + 1 - 1 over 2 cdot 8 - 1 + 1 over 2 cdot 8 + 1 + cdots = cr
& = 1 - sumlimits_1, le ,n 1 over 8n - 1 + sumlimits_1, le ,n 1 over 8n + 1 = cr
& = 1 - 1 over 8left( sumlimits_1, le ,n 1 over n - 1/8 - sumlimits_1, le ,n 1 over n + 1/8 right) = cr
& = 1 - 1 over 8left( sumnolimits_;n = 7/8^;infty 1 over n - sumnolimits_;n = 9/8^;infty 1 over n right) = cr
& = 1 - 1 over 8left( sumnolimits_;n = 7/8^;9/8 1 over n right) = cr
& = 1 - 1 over 8left( psi (9/8) - psi (7/8) right) = cr
& = 1 - 1 over 8left( psi left( 1 + 1/8 right) - psi left( 1 - 1/8 right) right) = cr
& = 1 - 1 over 8left( 1 over 1/8 + psi left( 1/8 right) - psi left( 1 - 1/8 right) right) = cr
& = 1 over 8left( psi left( 1 - 1/8 right) - psi left( 1/8 right) right) = cr
& = 1 over 8left( pi cot left( pi over 8 right) right) = cr
& = pi left( 1 + sqrt 2 right) over 8 cr
$$
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 $1/z$
$$
eqalign
& psi left( z right) = Delta _,z ^ - 1 left( 1 over z right) = sumnolimits_z 1 over z quad Rightarrow cr
& Rightarrow quad sumnolimits_;z = a^;b 1 over z = psi left( b right) - psi left( a right) cr
$$
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 $y$
$$
int_x = 0^,1 1 - x^,6 over 1 - x^,8 dx quad mathop
= limits^x^,8 = y quad 1 over 8int_y = 0^,1 1 - y^,3/4 over left( 1 - y right)y^,7/8 dy
$$
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
$$
eqalign
& 8 ;int_x = 0^,1 1 - x^,6 over 1 - x^,8 dx = int_y = 0^,1 1 - y^,3/4 over left( 1 - y right)y^,7/8 dy = cr
& = mathop lim limits_varepsilon , to ,0 left( int_y = 0^,1 left( 1 - y right)^varepsilon - 1 y^,1/8 - 1 dy
- int_y = 0^,1 left( 1 - y right)^varepsilon - 1 y^,,7/8 - 1 dy right) = cr
& = mathop lim limits_varepsilon , to ,0 left( rm Bleft( 1/8,varepsilon right) - rm Bleft( 7/8,varepsilon right) right) = cr
& = mathop lim limits_varepsilon , to ,0 left( Gamma left( 1/8 right)Gamma left( varepsilon right)
over Gamma left( 1/8 + varepsilon right)
- Gamma left( 7/8 right)Gamma left( varepsilon right) over Gamma left( 7/8 + varepsilon right) right) = cr
& = mathop lim limits_varepsilon , to ,0 left( Gamma left( 1/8 right)Gamma left( varepsilon right)
over psi left( 1/8 right)varepsilon ;Gamma left( 1/8 right)
+ Gamma left( 1/8 right) - Gamma left( 7/8 right)Gamma left( varepsilon right)
over psi left( 7/8 right)varepsilon ;Gamma left( 7/8 right) + Gamma left( 7/8 right) right) = cr
& = mathop lim limits_varepsilon , to ,0 left( Gamma left( varepsilon right) over psi left( 1/8 right)varepsilon ; + 1
- Gamma left( varepsilon right) over psi left( 7/8 right)varepsilon ; + 1 right) = cr
& = mathop lim limits_varepsilon , to ,0 Gamma left( varepsilon right)left( left( 1 - psi left( 1/8 right)varepsilon ; right)
- left( 1 - psi left( 7/8 right)varepsilon ; right) right) = cr
& = mathop lim limits_varepsilon , to ,0 varepsilon ,Gamma left( varepsilon right)left( psi left( 7/8 right) - psi left( 1/8 right) right) = cr
& = mathop lim limits_varepsilon , to ,0 ,Gamma left( 1 + varepsilon right)left( psi left( 7/8 right) - psi left( 1/8 right) right) = cr
& = psi left( 7/8 right) - psi left( 1/8 right) cr
$$
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
$$
psi (s + 1) = - gamma + int_0^1 1 - x^,s over 1 - xdx
$$

HINT: Notice that
$$fracx^4+x^2+1(x^2+1)(x^4+1)=frac14frac1x^2+xsqrt2+1+frac14frac1x^2-xsqrt2+1+frac12frac1x^2+1$$
Now we just have to evaluate the integrals
$$I_1=frac14int_0^1 fracdxx^2+xsqrt2+1$$
$$I_2=frac14int_0^1 fracdxx^2-xsqrt2+1$$
$$I_3=frac12int_0^1 fracdxx^2+1$$
and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?

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 $frac14left(2tan^-1x+sqrt2left(tan^-1left(sqrt2x+1right)-tan^-1left(1-sqrt2xright)right)right)$

Hint: your fraction is $(x^2+1)^-1+fracx^2x^4+1$, and $fracx^2x^4+1=fracax+bx^2-sqrt2x+1+frac-ax-bx^2+sqrt2x+1$ for some real numbers $a,b$.