Substitute [tex]y = x^{\frac{2n-1}2}[/tex] and [tex]dy = \frac{2n-1}2 x^{\frac{2n-3}2} \, dx[/tex]. Then the integral is
[tex]\displaystyle \int_0^\infty \frac{x^{\frac{2n-3}2} \tan^{-1}\left(x^{2n-1}\right)}{1 + x^{2n-1}} \, dx = \frac2{2n-1} \underbrace{\int_0^\infty \frac{\tan^{-1}\left(y^2\right)}{1 + y^2} \, dy}_I[/tex]
which is to say, the integral is now independent of n. Then
[tex]\displaystyle \sum_{n=1}^\infty (-1)^{n+1} \int_0^\infty \frac{x^{\frac{2n-3}2} \tan^{-1}\left(x^{2n-1}\right)}{1+x^{2n-1}} \, dx = I \sum_{n=1}^\infty \frac{2(-1)^{n+1}}{2n-1}[/tex]
Let's evaluate the sum. Recall that if |x| < 1, we have
[tex]\displaystyle \frac1{1-x} = \sum_{n=0}^\infty x^n[/tex]
which means
[tex]\displaystyle \frac1{1+x^2} = \frac1{1-(-x^2)} = \sum_{n=0}^\infty \left(-x^2\right)^n = \sum_{n=0}^\infty (-1)^n x^{2n}[/tex]
By the fundamental theorem of calculus, integrating both sides gives
[tex]\displaystyle \tan^{-1}(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^{2n-1}}{2n-1}[/tex]
As x approaches 1 from below, we have
[tex]\displaystyle \lim_{x\to1^-} \tan^{-1}(x) = \lim_{x\to1^-} \sum_{n=1}^\infty (-1)^{n+1} \frac{x^{2n-1}}{2n-1}[/tex]
[tex]\displaystyle \frac\pi4 = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1} \left(\lim\limits_{x\to1^-} x^{2n-1}\right) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}[/tex]
and so
[tex]\displaystyle \sum_{n=1}^\infty (-1)^{n+1} \int_0^\infty \frac{x^{\frac{2n-3}2} \tan^{-1}\left(x^{2n-1}\right)}{1+x^{2n-1}} \, dx = \frac\pi2 I[/tex]
Now compute the remaining integral. First split it at y = 1 :
[tex]\displaystyle I = \int_0^1 \frac{\tan^{-1}\left(y^2\right)}{1 + y^2} \, dy + \int_1^\infty \frac{\tan^{-1}\left(y^2\right)}{1 + y^2} \, dy[/tex]
In the second integral, notice that replacing [tex]z=\frac1y[/tex] and [tex]dy = -\frac{dz}{z^2}[/tex] yields
[tex]\displaystyle \int_1^\infty \frac{\tan^{-1}\left(y^2\right)}{1 + y^2} \, dy = - \int_1^0 \frac{\tan^{-1}\left(\frac1{z^2}\right)}{1 + \frac1{z^2}} \, \frac{dz}{z^2} = \int_0^1 \frac{\tan^{-1}\left(\frac1{z^2}\right)}{1+z^2} \, dz[/tex]
The inverse tangent function has the property
[tex]\tan^{-1}(x) = \dfrac\pi2 - \tan^{-1}\left(\dfrac1x\right)[/tex]
so it follows that
[tex]\displaystyle \int_0^1 \frac{\tan^{-1}\left(\frac1{z^2}\right)}{1+z^2} \, dz = \frac\pi2 \int_0^1 \frac{dz}{1+z^2} - \int_0^1 \frac{\tan^{-1}(z^2)}{1+z^2} \, dz[/tex]
and hence
[tex]\displaystyle I = \int_0^1 \frac{\tan^{-1}\left(y^2\right)}{1 + y^2} \, dy + \frac\pi2 \int_0^1 \frac{dz}{1+z^2} - \int_0^1 \frac{\tan^{-1}(z^2)}{1+z^2} \, dz[/tex]
[tex]\displaystyle I = \frac\pi2 \int_0^1 \frac{dz}{1+z^2}[/tex]
[tex]I = \dfrac\pi2 \times \dfrac\pi4 = \dfrac{\pi^2}8[/tex]
Then the original expression has an exact value of
[tex]\displaystyle \sum_{n=1}^\infty (-1)^{n+1} \int_0^\infty \frac{x^{\frac{2n-3}2} \tan^{-1}\left(x^{2n-1}\right)}{1+x^{2n-1}} \, dx = \frac\pi2 \times \frac{\pi^2}8 = \boxed{\frac{\pi^3}{16}}[/tex]
