1. ## Integrating Partial Fractions

Evalulate ∫

2. Originally Posted by evansf
Evalulate ∫
Hint:
$\frac{3x + 3}{x^3 - 1} = 3 \frac{x + 1}{(x - 1)(x^2 + x + 1)}$

So you are looking for partial fractions of the form
$\frac{x + 1}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + c}{x^2 + x + 1}$

-Dan

3. ## Yes

,,,

4. Originally Posted by evansf
Evalulate ∫
Let's just focus on $\frac{x+1}{(x-1)(x^2+x+1)}=\frac1{x^2+x+1}+\frac2{(x-1)(x^2+x+1)}.$

The first one is an arctangent. Let's see the second one:

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

This becomes to $\frac{1}{3}\left( {\frac{1}{{x - 1}} - \frac{{x + 2}}{{x^2 + x + 1}}} \right).$

Now locate the second term:

$\frac{{x + 2}}{{x^2 + x + 1}} = \frac{1}{2}\left( {\frac{{2x + 4}}{{x^2 + x + 1}}} \right) = \frac{1}{2}\left( {\frac{{2x + 1}}{{x^2 + x + 1}} + \frac{3}{{x^2 + x + 1}}} \right).$

This is to avoid partial fractions. (It's just another way.)