Rewrite what integral? Are you asked to find ?
, so
Now rewriting using partial fractions...
which gives
Solving these equations simultaneously gives , so that means
.
Can you go from here?
By factoring x^3 +1 as a sum of cubes, rewrite the integral in part a, then express
1/ (x^3 +1) as the sum of a power series and use it to prove the formula for pi.
n= [3(sq. rt 3)] / 4 , the sum of the series from n=0 to infinity [(-1) ^n] / 8^n [(2/3n+1) + (1/3n+2)]