Methods of Integration

(cod: P-74-59-2) Integration by Partial Fractions: Consider the sum below \[\dfrac{3}{x} + \dfrac{1}{x - 2} = \dfrac{3(x-2) + x}{x^2 - x} = \dfrac{4x - 6}{x^2 - x}.\] If we wanted to integrate the rational function \(g(x) = \dfrac{4x - 6}{x^2 - x}\), it would be interesting to work from the inverse perspective and consider it as the sum of simpler fractions that appear on the left side of the first equality.

To illustrate the process, let us take the rational function \[f(x) = \dfrac{-2x+1}{x^2 - 3x + 2}.\] The denominator \(Q(x) = x^2 - 3x + 2\) has two distinct real roots, namely, \(x = 1\) and \(x = 2\). Thus, \(Q(x)\) is reducible, meaning it can be factored as the product of two lower-degree polynomials (in this case, two polynomials of degree \(1\)): \[Q(x) = x^2 - 3x + 2 = (x-1)(x-2).\] We would then like to find constants \(A\) and \(B\) such that \[\dfrac{- 2x + 1}{x^2 - 3x + 2} = \dfrac{A}{x-1} + \dfrac{B}{x-2}.\] We can multiply both sides by \(x^2 - 3x + 2 = (x-1)(x-2)\), obtaining \begin{equation}\label{igual}- 2x+ 1 = A(x-2) + B(x-1).\end{equation} Grouping the terms on the right-hand side by degree, we get \[- 2x + 1 = (A+B)x + (-2A-B).\] Comparing the coefficients of the polynomials on both sides, we must have \[\left\{\begin{array}{l} A+B = -2 \\ -2A-B = 1 \end{array}\right.\] Solving the system, we find that \(A = 1\) and \(B=-3\). In this example, we could have proceeded differently to obtain the constants \(A\) and \(B\). Indeed, since we want the equality in (\ref{igual}) to hold for any value of \(x \in \mathbb{R}\), we can substitute specific values and easily find the values of \(A\) and \(B\). More precisely, setting \(x = 1\) in (\ref{igual}), we directly find that \(A = 1\), while setting \(x=2\), we conclude that \(B = -3\). Therefore, we have that \begin{eqnarray*} & & \int \dfrac{-2x + 1}{x^2 - 3x + 2} \, dx \\ & & \\ &=& \int \left(\dfrac{1}{x-1} - \dfrac{3}{x-2} \right) \, dx \\ & & \\ &=& \ln|x-1| -3\ln|x-2| + C. \end{eqnarray*} This example illustrates the method of partial fractions. It is a method for integrating rational functions that involves expressing a function of this class (which is a ratio of polynomials) as the sum of simpler fractions (called partial fractions), which we can integrate. We then proceed to describe the method in a more systematic way.

We say that a rational function is proper if the degree of the numerator is less than the degree of the denominator. Otherwise, we say that the rational function is improper. It is always possible to write an improper rational function \(h(x) = \dfrac{P(x)}{Q(x)}\) as the sum of a polynomial and a proper rational function by applying the division algorithm. Indeed, if the quotient of the division of \(P(x)\) by \(Q(x)\) is \(A(x)\), with remainder \(R(x)\), we have \[P(x) = Q(x)A(x) + R(x),\] where the degree of \(R(x)\) is less than the degree of \(Q(x)\). Dividing the above equality by \(Q(x)\), we then obtain \[\dfrac{P(x)}{Q(x)} = A(x) + \dfrac{R(x)}{Q(x)},\] where the ratio on the right-hand side is a proper rational function.

We will describe the method considering proper rational functions, applying the above procedure if the function is improper. It is possible to demonstrate that the decomposition strategy presented below always works, in the sense that it is always possible to determine the constants and, finally, integrate all the partial fractions that arise.

From now on, let us consider a proper rational function \(f(x) = \dfrac{P(x)}{Q(x)},\) where the numerator and denominator are coprime, that is, they have no common factors. The decomposition of \(f\) as a sum of partial fractions depends on the factorization of \(Q(x)\) into irreducible factors. We have the following fact:

Every polynomial with real coefficients admits a factorization as a product of irreducible factors of degree 1 or 2 (although, in practice, it is difficult to exhibit such a factorization).

We now briefly explain the above fact. The Fundamental Theorem of Algebra guarantees that \(Q(x)\) has \(n\) roots (not necessarily distinct) in the set of complex numbers, some of which may be real. Starting from the \(n\) roots \(z_1, z_2, \dots,z_n\) of \(Q(x)\) in \(\mathbb{C}\), we first verify that \(Q(x)\) admits a decomposition (using complex numbers) in the form \begin{equation*}\label{dec1}Q(x) = c_n(x - z_1) (x-z_2)\cdots (x-z_n), \end{equation*} where \(c_n \in \mathbb{R}\) is the leading coefficient of \(Q(x)\) (note that we can consider \(c_n = 1\) after a simple algebraic manipulation in \(f(x)\)). Since \(Q(x)\) has real coefficients, it is possible to ensure that, for each non-real complex root \(z\) of \(Q(x)\), its conjugate \(\overline{z}\) will also be a root, simply by using that \[Q(\overline{z}) = \overline{Q(z)} = \overline{0} = 0.\] Note, then, that the product of the factors \((x - z)(x-\overline{z})\) gives rise to the quadratic polynomial, \[x^2 - \underbrace{(z+\overline{z})}_{\in \mathbb{R}} x + \underbrace{z\overline{z}}_{\in \mathbb{R}},\] which has real coefficients and whose roots are precisely \(z\) and \(\overline{z}\) (negative discriminant). Thus, starting from the above decomposition, we can obtain the mentioned factorization, where each real root of \(Q(x)\) gives rise to a degree 1 factor, while non-real roots appear in pairs (conjugates), giving rise to an irreducible quadratic factor.

Grouping the repeated factors (roots with multiplicity greater than 1), we can consider the factorization \[Q(x) = q_1(x)q_2(x) \cdots q_m(x),\] where a block \(q_j(x)\) can take the form \begin{equation}\label{bloco1}(x - r)^{l} \end{equation} or \begin{equation}\label{bloco2}(x^2 + bx + c)^{k}, \end{equation} with the first case associated with a real root \(r\) of multiplicity \(l\), and the second associated with a pair \(z\) and \(\overline{z}\) of conjugate complex roots, each with multiplicity \(k\).

From these blocks, we establish the decomposition of the proper rational function \(f(x) = \dfrac{P(x)}{Q(x)}\) as a sum of partial fractions. For each block of the form (\ref{bloco1}), we add \(l\) partial fractions to the decomposition: \[\dfrac{A_1}{x-r} + \dfrac{A_2}{(x-r)^2} + \cdots + \dfrac{A_l}{(x-r)^l},\] where \(A_1,A_2,\dots,A_l\) are constants to be determined. On the other hand, for each block of the form (\ref{bloco2}), we add \(k\) partial fractions to the decomposition: \[\scriptsize{\dfrac{B_1 x + C_1 }{x^2 + bx + c} + \dfrac{B_2 x + C_2 }{(x^2 + bx + c)^2} + \cdots + \dfrac{B_k x + C_k }{(x^2 + bx + c)^k},}\] where \(B_j\) and \(C_j\), with \(j = 1,\dots,k\), are constants to be determined. The decomposition into partial fractions is given by the sum of all fractions, considering all blocks.

To make it clearer, consider an example where the denominator factors as \begin{eqnarray*}Q(x) &=& (x-1)(x-1)(x^2+4) \\ & & \\ &=& (x-1)^2 (x^2+4). \end{eqnarray*} The degree \(1\) factor appears twice (\(x = 1\) is a double root), while \(x^2+4\) is an irreducible quadratic factor (whose roots are \(z = 2i\) and \(\overline{z} = - 2i\)). The decomposition into partial fractions of \(f\) would then be \[\dfrac{P(x)}{Q(x)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{Cx+D}{x^2 + 4}.\] Multiplying both sides by \(Q(x)\) and comparing the coefficients on the left and right sides, we can calculate the constants \(A\), \(B\), \(C\), and \(D\), and then calculate the integral of \(f\).

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As explained above, the decomposition into partial fractions of the proper rational function \(f(x) = \dfrac{1}{x(x^2+4)^2}\) is given by \[\dfrac{1}{x(x^2+4)^2} = \dfrac{A}{x} + \dfrac{Bx + C}{x^2+4} + \dfrac{Dx + E}{(x^2+4)^2} \] Consider the proper rational function \[f(x) = \dfrac{3x^2 - x + 2}{x^3 + x}.\] We can factor the denominator, obtaining \[x^3 + x = x(x^2+1),\] noting that the factor \((x^2 + 1)\) is irreducible. Following the method described above, we must decompose \(f(x)\) as \[\dfrac{3x^2 - x + 2}{x^3 + x} = \dfrac{A}{x} + \dfrac{Bx + C}{x^2 + 1}.\] Consider the proper rational function \[f(x) = \dfrac{3x^2 + 2x - 3}{x^3 - x}.\] Using the method described above, its decomposition into partial fractions is given by: \[\dfrac{3x^2 + 2x - 3}{x^3 - x} = \dfrac{A}{x} + \dfrac{B}{x-1} + \dfrac{C}{x+1} \] The rational function \[h(x) = \dfrac{2x^3 - 2x^2 + 1}{x^2 - x}\] is improper. After performing the division algorithm, we can write it in the form \[h(x) = 2x + \dfrac{1}{x^2 - x},\] where the ratio on the right-hand side is a proper rational function. Consider the proper rational function \(f(x) = \dfrac{1}{x(x^2-x)}\). The decomposition into partial fractions of \(f\) is given by \[\dfrac{1}{x(x^2-x)} = \dfrac{A}{x} + \dfrac{Bx + C}{x^2 - x}.\]