Integration of rational function whose denominator has distinct linear factors

Outline of method

Reduction to the case where the numerator has smaller degree than the denominator

For further information, refer: converting a rational function from improper fraction to mixed fraction form

To get to the situation where the numerator has smaller degree than the denominator, we perform a Euclidean division and hence rewrite the rational function as the sum of a polynomial and a rational function that is in proper fraction form, i.e., the numerator has smaller degree than the denominator. The polynomial summand is integrated termwise using the integration rule for power functions. We are thus reduced to handling the proper fraction. But remember to add back the antiderivative for the polynomial to your final answer!

Reduction to the monic denominator case

If the leading coefficient (i.e., the coefficient on the highest degree term) in the denominator is not 1, the leading coefficient can be pulled out as a constant factor from the denominator and hence out of the integration. We can thus carry out an integration with a leading coefficient of 1 for the denominator polynomial (such a polynomial is termed a monic polynomial). But remember to keep that constant on the outside and multiply it to get your final answer!

Integration in the reduced case

We have the following formula for the integration, where $r(x)$ has degree less than $n$, and $\alpha _{1},\alpha _{2},\dots ,\alpha _{n}$ are pairwise distinct real numbers:

$\!\int {\frac {r(x)\,dx}{(x-\alpha _{1})(x-\alpha _{2})\dots (x-\alpha _{n})}}=\sum _{i=1}^{n}{\frac {r(\alpha _{i})}{\prod _{1\leq j\leq n,j\neq i}(\alpha _{i}-\alpha _{j})}}\ln |x-\alpha _{i}|+C$

The formula is a little complicated to understand, so we write it down explicitly for the first few values of $n$.

$n$	What does $r(x)$ look like?	Integration formula
1	constant function	$\!\int {\frac {r(x)\,dx}{x-\alpha _{1}}}=r(\alpha _{1})\ln |x-\alpha _{1}|+C$. Note that the formula denominator, being an empty product, becomes 1.
2	constant or linear function	$\!\int {\frac {r(x)\,dx}{(x-\alpha _{1})(x-\alpha _{2})}}={\frac {r(\alpha _{1})}{\alpha _{1}-\alpha _{2}}}\ln |x-\alpha _{1}|+{\frac {r(\alpha _{2})}{\alpha _{2}-\alpha _{1}}}\ln |x-\alpha _{2}|+C$
3	constant, linear, or quadratic function	$\!\int {\frac {r(x)\,dx}{(x-\alpha _{1})(x-\alpha _{2})(x-\alpha _{3})}}={\frac {r(\alpha _{1})}{(\alpha _{1}-\alpha _{2})(\alpha _{1}-\alpha _{3})}}\ln |x-\alpha _{1}|+{\frac {r(\alpha _{2})}{(\alpha _{2}-\alpha _{1})(\alpha _{2}-\alpha _{3})}}\ln |x-\alpha _{2}|+{\frac {r(\alpha _{3})}{(\alpha _{3}-\alpha _{1})(\alpha _{3}-\alpha _{2})}}\ln |x-\alpha _{3}|+C$

Definite integration

For simplicity, suppose the rational function, written in reduced form, i.e., with no common factor between the numerator and denominator, has denominator:

$(x-\alpha _{1})(x-\alpha _{2})\dots (x-\alpha _{n})$

with $\alpha _{1}<\alpha _{2}<\dots <\alpha _{n}$ (the ordering assumption is not necessary for the integration formulas above to hold, but helps simplify the discussion below).

Then, the domain of the rational function is the complement of the subset $\{\alpha _{1},\alpha _{2},\dots ,\alpha _{n}\}$ of $\mathbb {R}$. This is a union of $n+1$ pairwise disjoint open intervals:

$(-\infty ,\alpha _{1}),(\alpha _{1},\alpha _{2}),(\alpha _{2},\alpha _{3}),\dots ,(\alpha _{n-1},\alpha _{n}),(\alpha _{n},\infty )$

In order for a definite integration to make sense, the two endpoints of the integration must both lie within the same open interval. In other words, we can integrate over any closed interval that lies inside one of these $n+1$ open intervals.

Further, if we are interested in an antiderivative on the whole domain, then, in the antiderivative expression above, the constant value is constant within each of the $n+1$ intervals but can differ between the intervals.

Finally, note that in any fixed choice of interval, all the signs of the expressions $x-\alpha _{i}$ are determinate (i.e., each sign is either positive throughout the interval or negative throughout the interval). We can thus replace the absolute value by either $x-\alpha _{i}$ or $-(x-\alpha _{i})$ for each $i$ depending on the choice of $i$ and the interval. Specifically:

• In the interval $(-\infty ,\alpha _{1})$, all the $x-\alpha _{i}$s are negative, so $\ln |x-\alpha _{i}|=\ln(\alpha _{i}-x)$.
• In the interval $(\alpha _{n},\infty )$, all the $x-\alpha _{i}$s are positive, so $\ln |x-\alpha _{i}|=\ln(x-\alpha _{i})$.
• In the interval $(\alpha _{j},\alpha _{j+1})$, $x-\alpha _{i}$ is positive for $1\leq i\leq j$ (so $\ln |x-\alpha _{i}|=\ln(x-\alpha _{i})$ for these $i$s) and negative for $j+1\leq i\leq n$ (so $\ln |x-\alpha _{i}|=\ln(\alpha _{i}-x)$ for these $i$s).