Rational functions: the development into partial fractions.

    

Recall that:

We decompose a rational fraction into a sum of partial fractions according to the following requirements:

Example 8.2.10   Let $ F(x)=\int \frac {1}{1-x^2} \; dx$ , for $ -1 < x < 1$ . We have: $ \forall x \in (-1,1), \; \frac {1}{1-x^2}= \frac {1}{(1-x)(1+x)}
= \frac 12 \cdot \frac {1}{1+x} + \frac 12 \cdot \frac {1}{1-x}$ . Therefore:

$\displaystyle \forall x \in (-1,1), \; F(x)$ $\displaystyle = \frac 12 \int \frac {1}{1+x} \; dx + \frac 12 \int \frac {1}{1-x} \; dx$    
$\displaystyle \quad$ $\displaystyle =\frac 12 \ln \vert 1+x\vert - \frac 12 \ln \vert 1-x\vert +C$    
$\displaystyle \quad$ $\displaystyle = \frac 12 \ln \left\vert \frac {1+x}{1-x} \right\vert +C.$    

Example 8.2.11   Let $ F(x)=\int \frac {2x}{x^3-1} \; dx$ , for $ x>1$ . We have:

$\displaystyle \forall x \in (1,+\infty ), \; \frac {2x}{x^3-1}$ $\displaystyle = -\frac 23 \cdot \frac {x-1}{x^2+x+1} + \frac 23 \cdot \frac {1}{x-1}$    
$\displaystyle \quad$ $\displaystyle -\frac 13 \frac {2x+1}{x^2+x+1} + \frac {1}{x^2+x+1} + \frac 23 \cdot \frac {1}{x-1}$    

We compute separately the integral of each partial fraction; the first one and the third one use the techniques from  2.1, the second one requires the techniques of 2.2 (i.e. a substitution). Finally, we have:

$\displaystyle \forall x \in (1,+\infty ),$    
$\displaystyle F(x)$ $\displaystyle = -\frac 13 \ln \vert x^2+x+1\vert + \frac {2 \sqrt{3}}{3} \arctan \left( \frac {\sqrt{3}}{3} (2x+1) \right) + \frac 23 \ln \vert x-1\vert +C$    
$\displaystyle \quad$ $\displaystyle = \frac 13 \ln \frac {(x-1)^2}{x^2+x+1} + \frac {2 \sqrt{3}}{3} \arctan \left( \frac {\sqrt{3}}{3} (2x+1) \right) +C.$    

Noah Dana-Picard 2007-12-28