Calculation:

Consider the given integral equation is,

  ∫012dxx2−x+1=π33

Solve the left hand side as,

  ∫012dxx2−x+1=∫012dxx2−x+14−14+1=∫012dx(x−12)2+34=[132tan−1(x−12)32]012=23[tan−10−tan−1(1−13)]

Solve further as,

  ∫012dxx2−x+1=π33

Hence, proved.

Given:

The given formula for π is,

  π=334∑n=0∞(−1)n8n(23n+1+13n+2)

Calculation:

Consider the expression x2+1=(x+1)(x2−x+1)

Then,

  1x2−x+1=1+x1+x3⋅1+x1−(−x3)=(1+x)∑(−x3)n=(1+x)∑(−1)nx3n=∑n=0∞(−1)nx3n+∑n=0∞(−1)nx3n+1 for |x|<1

Therefore,

  ∫dxx2−x+1=∫|∑n=0∞(−1)nx3n+∑n=0∞(−1)nx3n+1 for |x|<1|dx=|∑n=0∞(−1)nx3n+13n+1+∑n=0∞(−1)nx3n+23n+2          for |x|<1|

Then,

  ∫0121x2−x+1dx=|∑n=0∞(−1)n(12)3n+13n+1+∑n=0∞(−1)n(12)3n+23n+2|=∑n=0∞(−1)n(12)3n+13n+1+∑n=0∞(−1)n(12)3n+23n+2=∑n=0∞(−1)n4⋅23n[23n+1+13n+2]=∑n=0∞(−1)n4⋅83n[23n+1+13n+2]

From part (a) ∫012dxx2−x+1=π33 then,

  π33=334∑n=0∞(−1)n8n[23n+1+13n+2]

Hence, proved.