Rational functions: the development into partial fractions.

    

Recall that:

• A rational fraction is an expression of the form $r(x)=\frac {f(x)}{g(x)}$ , where $f(x)$ and $g(x)$ are two polynomials.
• If $\deg f(x) < \deg g(x)$ , the rational fraction is proper; otherwise it is improper .
• An improper rational fraction is the sum of a polynomial and a proper rational fraction.

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

• If the denominator of $r(x)$ contains a factor $ax+b$ of multiplicity 1, then there is a single fraction of the form $\frac {\alpha}{ax+b}$ in the desired developement ($\alpha$ is a constant).
• If the denominator of $r(x)$ contains a factor $ax+b$ of multiplicity $m$ , then the decomposition of $r(x)$ into partial fractions contains a sum of fractions of the form $\underset{k=1}{\overset{m}{\sum}} \frac {\alpha_k }{(ax+b)^k}$ , where the $\alpha_k$ are constants.
• If the denominator of $r(x)$ contains an irreducible factor $ax^2+bx+c$ of multiplicity 1, then there is a single fraction of the form $\frac {\alpha x + \beta}{ax^2+bx+c}$ in the desired developement ($\alpha$ and $\beta$ are constants).
• If the denominator of $r(x)$ contains an irreducible factor $ax^2+bx+c$ of multiplicity $m$ , then the decomposition of $r(x)$ into partial fractions contains a sum of fractions of the form $\underset{k=1}{\overset{m}{\sum}} \frac {\alpha_k x + \beta_k}{(ax^2+bx+c)^k}$ , where the $\alpha_k$ and $\beta_k$ are constants.

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:

$$\forall x \in (-1,1), \; F(x) = \frac 12 \int \frac {1}{1+x} \; dx + \frac 12 \int \frac {1}{1-x} \; dx$$

$$\quad =\frac 12 \ln \vert 1+x\vert - \frac 12 \ln \vert 1-x\vert +C$$

$$\quad = \frac 12 \ln \left\vert \frac {1+x}{1-x} \right\vert +C.$$

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

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

$$\quad -\frac 13 \frac {2x+1}{x^2+x+1} + \frac {1}{x^2+x+1} + \frac 23 \cdot \frac {1}{x-1}$$

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:

$$\forall x \in (1,+\infty ),$$

$$F(x) = -\frac 13 \ln \vert x^2+x+1\vert + \frac {2 \sqrt{3}}{3} \arctan \left( \frac {\sqrt{3}}{3} (2x+1) \right) + \frac 23 \ln \vert x-1\vert +C$$

$$\quad = \frac 13 \ln \frac {(x-1)^2}{x^2+x+1} + \frac {2 \sqrt{3}}{3} \arctan \left( \frac {\sqrt{3}}{3} (2x+1) \right) +C.$$