130-131. Integration of rational functions

130. Rational Functions.

After integrating polynomials it is natural to turn our attention next to rational functions. Let us suppose $R (x)$ to be any rational function expressed in the standard form of § 117, viz. as the sum of a polynomial $Π (x)$ and a number of terms of the form $A / (x - α)^{p}$ .

We can at once write down the integrals of the polynomial and of all the other terms except those for which $p = 1$ , since $\int \frac{A}{(x - α)^{p}} d x = - \frac{A}{p - 1} \frac{1}{(x - α)^{p - 1}},$ whether $α$ be real or complex ( § 117).

The terms for which $p = 1$ present rather more difficulty. It follows immediately from Theorem (6) of § 113 that $\begin{matrix} (3) & \int F^{'} {f (x)} f^{'} (x) d x = F {f (x)} . \end{matrix}$ In particular, if we take $f (x) = a x + b$ , where $a$ and $b$ are real, and write $ϕ (x)$ for $F (x)$ and $ψ (x)$ for $F^{'} (x)$ , so that $ϕ (x)$ is an integral of $ψ (x)$ , we obtain $\begin{matrix} (4) & \int ψ (a x + b) d x = \frac{1}{a} ϕ (a x + b) . \end{matrix}$

Thus, for example, $\int \frac{d x}{a x + b} = \frac{1}{a} \log | a x + b |,$ and in particular, if $α$ is real, $\int \frac{d x}{x - α} = \log | x - α | .$ We can therefore write down the integrals of all the terms in $R (x)$ for which $p = 1$ and $α$ is real. There remain the terms for which $p = 1$ and $α$ is complex.

In order to deal with these we shall introduce a restrictive hypothesis, viz. that all the coefficients in $R (x)$ are real. Then if $α = γ + δ i$ is a root of $Q (x) = 0$ , of multiplicity $m$ , so is its conjugate $\bar{α} = γ - δ i$ ; and if a partial fraction $A_{p} / (x - α)^{p}$ occurs in the expression of $R (x)$ , so does ${\bar{A}}_{p} / (x - \bar{α})^{p}$ , where ${\bar{A}}_{p}$ is conjugate to $A_{p}$ . This follows from the nature of the algebraical processes by means of which the partial fractions can be found, and which are explained at length in treatises on Algebra.¹

Thus, if a term $(λ + μ i) / (x - γ - δ i)$ occurs in the expression of $R (x)$ in partial fractions, so will a term $(λ - μ i) / (x - γ + δ i)$ ; and the sum of these two terms is $\frac{2 {λ (x - γ) - μ δ}}{(x - γ)^{2} + δ^{2}} .$ This fraction is in reality the most general fraction of the form $\frac{A x + B}{a x^{2} + 2 b x + c},$ where $b^{2} < a c$ . The reader will easily verify the equivalence of the two forms, the formulae which express $λ$ , $μ$ , $γ$ , $δ$ in terms of $A$ , $B$ , $a$ , $b$ , $c$ being $λ = A / 2 a, μ = - D / (2 a \sqrt{Δ}), γ = - b / a, δ = \sqrt{Δ} / a,$ where $Δ = a c - b^{2}$ , and $D = a B - b A$ .

If in (3) we suppose $F {f (x)}$ to be $\log | f (x) |$ , we obtain $\begin{matrix} (5) & \int \frac{f^{'} (x)}{f (x)} d x = \log | f (x) |; \end{matrix}$ and if we further suppose that $f (x) = (x - λ)^{2} + μ^{2}$ , we obtain $\int \frac{2 (x - λ)}{(x - λ)^{2} + μ^{2}} d x = \log {(x - λ)^{2} + μ^{2}} .$ And, in virtue of the equations (6) of § 128 and (4) above, we have $\int \frac{- 2 δ μ}{(x - λ)^{2} + μ^{2}} d x = - 2 δ \arctan (\frac{x - λ}{μ}) .$

These two formulae enable us to integrate the sum of the two terms which we have been considering in the expression of $R (x)$ ; and we are thus enabled to write down the integral of any real rational function, if all the factors of its denominator can be determined. The integral of any such function is composed of

the sum of a polynomial, a number of rational functions of the type $- \frac{A}{p - 1} \frac{1}{(x - α)^{p - 1}},$ a number of logarithmic functions, and a number of inverse tangents.

It only remains to add that if $α$ is complex then the rational function just written always occurs in conjunction with another in which $A$ and $α$ are replaced by the complex numbers conjugate to them, and that the sum of the two functions is a real rational function.

Example XLVIII

1. Prove that $\int \frac{A x + B}{a x^{2} + 2 b x + c} d x = \frac{A}{2 a} \log | X | + \frac{D}{2 a \sqrt{- Δ}} \log | \frac{a x + b - \sqrt{- Δ}}{a x + b + \sqrt{- Δ}} |$ (where $X = a x^{2} + b x + c$ ) if $Δ < 0$ , and $\int \frac{A x + B}{a x^{2} + 2 b x + c} d x = \frac{A}{2 a} \log | X | + \frac{D}{2 a \sqrt{Δ}} \arctan (\frac{a x + b}{\sqrt{Δ}})$ if $Δ > 0$ , $Δ$ and $D$ having the same meanings as above.

2. In the particular case in which $a c = b^{2}$ the integral is $- \frac{D}{a (a x + b)} + \frac{A}{a} \log | a x + b | .$

3. Show that if the roots of $Q (x) = 0$ are all real and distinct, and $P (x)$ is of lower degree than $Q (x)$ , then $\int R (x) d x = \sum \frac{P (α)}{Q^{'} (α)} \log | x - α |,$ the summation applying to all the roots $α$ of $Q (x) = 0$ .

[The form of the fraction corresponding to

α

may be deduced from the facts that

\frac{Q (x)}{x - α} \to Q^{'} (α), (x - α) R (x) \to \frac{P (α)}{Q^{'} (α)},

x \to α

4. If all the roots of $Q (x)$ are real and $α$ is a double root, the other roots being simple roots, and $P (x)$ is of lower degree than $Q (x)$ , then the integral is $A / (x - α) + A^{'} \log | x - α | + \sum B \log | x - β |$ , where $A = - \frac{2 P (α)}{Q ” (α)}, A^{'} = \frac{2 {3 P^{'} (α) Q ” (α) - P (a) Q ”^{'} (α)}}{3 {Q ” (α)}^{2}}, B = \frac{P (β)}{Q^{'} (β)},$ and the summation applies to all roots $β$ of $Q (x) = 0$ other than $α$ .

5. Calculate $\int \frac{d x}{{(x - 1) (x^{2} + 1)}^{2}} .$

[The expression in partial fractions is

\frac{1}{4 (x - 1)^{2}} - \frac{1}{2 (x - 1)} - \frac{i}{8 (x - i)^{2}} + \frac{2 - i}{8 (x - i)} + \frac{i}{8 (x + i)^{2}} + \frac{2 + i}{8 (x + i)},

and the integral is

- \frac{1}{4 (x - 1)} - \frac{1}{4 (x^{2} + 1)} - \frac{1}{2} \log | x - 1 | + \frac{1}{4} \log (x^{2} + 1) + \frac{1}{4} \arctan x .]

6. Integrate $\begin{matrix} \frac{x}{(x - a) (x - b) (x - c)}, \frac{x}{(x - a)^{2} (x - b)}, \frac{x}{(x - a)^{2} (x - b)^{2}}, \frac{x}{(x - a)^{3}}, \\ \frac{x}{(x^{2} + a^{2}) (x^{2} + b^{2})}, \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b)^{2}}, \frac{x^{2} - a^{2}}{x^{2} (x^{2} + a^{2})}, \frac{x^{2} - a^{2}}{x (x^{2} + a^{2})^{2}} . \end{matrix}$

7. Prove the formulae: $\begin{aligned} \int \frac{d x}{1 + x^{4}} = \frac{1}{4 \sqrt{2}} { & \log (\frac{1 + x \sqrt{2} + x^{2}}{1 - x \sqrt{2} + x^{2}}) + 2 \arctan (\frac{x \sqrt{2}}{1 - x^{2}})}, \\ \int \frac{x^{2} d x}{1 + x^{4}} = \frac{1}{4 \sqrt{2}} { & - \log (\frac{1 + x \sqrt{2} + x^{2}}{1 - x \sqrt{2} + x^{2}}) + 2 \arctan (\frac{x \sqrt{2}}{1 - x^{2}})}, \\ \int \frac{d x}{1 + x^{2} + x^{4}} = \frac{1}{4 \sqrt{3}} { & \sqrt{3} \log (\frac{1 + x + x^{2}}{1 - x + x^{2}}) + 2 \arctan (\frac{x \sqrt{3}}{1 - x^{2}})} . \end{aligned}$

131. Note on the practical integration of rational functions.

The analysis of § 130 gives us a general method by which we can find the integral of any real rational function $R (x)$ , provided we can solve the equation $Q (x) = 0$ . In simple cases (as in Ex. 5 above) the application of the method is fairly simple. In more complicated cases the labour involved is sometimes prohibitive, and other devices have to be used. It is not part of the purpose of this book to go into practical problems of integration in detail. The reader who desires fuller information may be referred to Goursat’s Cours d’Analyse, second ed., vol. i, pp. 246 et seq., Bertrand’s Calcul Intégral, and Dr Bromwich’s tract Elementary Integrals (Bowes and Bowes, 1911).

If the equation $Q (x) = 0$ cannot be solved algebraically, then the method of partial fractions naturally fails and recourse must be had to other methods.²

See, for example, Chrystal’s Algebra, vol. i, pp. 151–9.↩︎
See the author’s tract “The integration of functions of a single variable” (Cambridge Tracts in Mathematics, No. 2, second edition, 1915). This does not often happen in practice.↩︎

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129. Integration of polynomials

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132-139. Integration of algebraical functions. Integration by rationalisation. Integration by parts

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