Another Derivation of the Bailey-Borwein-Plouffe Series

We know the following integral equals pi:

Now add and subtract the integral

I2 := (∫_0^1)(( 4 u)/((u^2) + 
                1)) \[DifferentialD]u

So we have I1 - I2, where

I1 := (∫_0^1)(( 4 (u + 1))/((u^2) + 
                1)) \[DifferentialD]u

I1-I2//FullSimplify

In I1, make the change of variable u = 1 - y.
Then 1 + u = 2 - y,   u = -dy
1 + u^2 = 2 - 2 y + y^2
As u goes from 0 to 1, y goes from 1 to 0,
so we get the same limits of integration with an extra minus sign.

I1New := (∫_0^1)(( 
                4 (2 - y))/((y^2) - 2 y + 
                2)) \[DifferentialD]y

In I2, make the change of variable u^2 = 1 - y^2.
Then u du = - y dy, and 1 + u^2 = 2 - y^2
As u goes from 0 to 1, y goes from 1 to 0,
so we get the same limits of integration with an extra minus sign.

I2New := (∫_0^1)(( 4 y)/(2 - 
                (y^2))) \[DifferentialD]y

I1New-I2New//FullSimplify

Now just combine the integrands in I1New and I2New

F  =  
      4 ( (2 - y)/(y^2  +  2  -  2  y)  -  
              y/(2 - y^2)) // Together

This is the original Bailey-Borwein-Plouffe integral

(∫_0^1) F \[DifferentialD]y // FullSimplify