1. ## Completing the Square

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)]

2. Rewrite what integral? Are you asked to find $\displaystyle \int{\frac{1}{x^3 + 1}\,dx}$?

$\displaystyle (x^3 + 1) = (x + 1)(x^2 - x + 1)$, so

$\displaystyle \frac{1}{x^3 + 1} = \frac{1}{(x + 1)(x^2 - x + 1)}$

Now rewriting using partial fractions...

$\displaystyle \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1} = \frac{1}{(x + 1)(x^2 - x + 1)}$

$\displaystyle \frac{A(x^2 - x + 1) + (Bx + C)(x + 1)}{(x + 1)(x^2 - x + 1)} = \frac{1}{x^2 - x + 1}$

$\displaystyle A(x^2 - x + 1) + (Bx + C)(x + 1) = 1$

$\displaystyle Ax^2 - Ax + A + Bx^2 + Bx + Cx + C = 1$

$\displaystyle (A + B)x^2 + (-A + B + C)x + A + C = 0x^2 + 0x + 1$

which gives

$\displaystyle A + B = 0$
$\displaystyle -A + B + C = 0$
$\displaystyle A + C = 1$

Solving these equations simultaneously gives $\displaystyle A = \frac{1}{3}, B = -\frac{1}{3}, C = \frac{2}{3}$, so that means

$\displaystyle \frac{1}{(x + 1)(x^2 - x + 1)} = \frac{1}{3(x + 1)} + \frac{(-x + 2)}{3(x^2 - x + 1)}$.

Can you go from here?