the formula or in the special case when xi is a simple root, when Over the reals Partial fractions are used in real-variable integral calculus to find real-valued antiderivatives of rational functions. Partial fraction decomposition of real rational functions is also used to find their Inverse Laplace transforms. For applications of partial fraction decomposition over the reals, see Application to symbolic integration, above Partial fractions in Laplace transforms General result Let f(x) be any rational function over the real numbers. In other words, suppose there exist real polynomials functions p(x) and q(x) ≠ 0, such that By dividing both the numerator and the denominator by the leading coefficient of q(x), we may assume without loss of generality that q(x) is monic. By the fundamental theorem of algebra, we can write where a1,..., am, b1,..., bn, c1,..., cn are real numbers with bi2 − 4ci < 0, and j1,..., jm, k1,..., kn are positive integers. The terms (x − ai) are the linear factors of q(x) which correspond to real roots of q(x), and the terms (xi2 + bix + ci) are the irreducible quadratic factors of q(x) which correspond to pairs of complex conjugate roots of q(x). Then the partial fraction decomposition of f(x) is the following: Here, P(x) is a (possibly zero) polynomial, and the Air, Bir, and Cir are real constants. There are a number of ways the constants can be found. The most straightforward method is to multiply through by the common denominator q(x). We then obtain an equation of polynomials whose left-hand side is simply p(x) and whose right-hand side has coefficients which are linear expressions of the constants Air, Bir, and Cir. Since two polynomials are equal if and only if their corresponding coefficients are equal, we can equate the coefficients of like terms. In this way, a system of linear equations is obtained which always has a unique solution. This solution can be found using any of the standard methods of linear algebra. It can also be found with limits (see Example 5). Examples Example 1 Here, the denominator splits into two distinct linear factors: so we have the partial fraction decomposition Multiplying through by the denominator on the left-hand side gives us the polynomial identity Substituting x = −3 into this equation gives A = −1/4, and substituting x = 1 gives B = 1/4, so that Example 2 After long division, we