Partial fractions for a factorable denominator, integrals from easy to advanced lesson

The integral of $1/(x^2 - 1)$ from $2$ to $3$ does not match a basic antiderivative as it stands, but its denominator factors as $(x - 1)(x + 1)$ . A fraction over a product of distinct linear factors can be written as a sum of simpler fractions, one over each factor, and each of those integrates to a logarithm. This page shows how to find the coefficients exactly, why the zeros of the denominator have to lie outside the interval of integration, and how the pieces combine to the value $\tfrac12\ln\tfrac32$ .

Key terms

Rational function

Definition A rational function is a quotient $R(x) = P(x)/Q(x)$ of polynomials $P$ and $Q$ with real coefficients, where $Q$ is not the zero polynomial. It is defined at every real $x$ with $Q(x) \ne 0$ and is continuous there. It is proper when $\deg P < \deg Q$ . When $\deg P \ge \deg Q$ , polynomial long division gives polynomials $S$ and $T$ with $P = SQ + T$ , where $T = 0$ or $\deg T < \deg Q$ , so that $P/Q = S + T/Q$ is a polynomial plus a proper rational function.

In plain words A rational function is a fraction of polynomials. Its domain leaves out the zeros of the denominator, and partial fractions apply only once the numerator has smaller degree than the denominator, which long division arranges.

Example The integrand $1/(x^2 - 1)$ is proper, with a numerator of degree $0$ and a denominator of degree $2$ , and it is defined for every real $x$ except $1$ and $-1$ . The function $x^3/(x^2 - 1)$ is not proper. Division gives $x^3 = x(x^2 - 1) + x$ , so

\frac{x^3}{x^2 - 1} = x + \frac{x}{x^2 - 1},

and only the remainder $x/(x^2 - 1)$ needs to be split into partial fractions.

Partial fraction decomposition

Definition Let $P/Q$ be a proper rational function with real coefficients, and factor $Q$ over the reals as a constant times powers of distinct linear factors $x - r$ and of distinct irreducible quadratic factors $x^2 + bx + c$ , those with $b^2 - 4c < 0$ . Then $P/Q$ is a sum of the following terms, with real constants that are uniquely determined. A factor $(x - r)^m$ contributes

\begin{aligned} &\frac{A_1}{x - r} + \frac{A_2}{(x - r)^2} \\[0.4em] &\quad + \cdots + \frac{A_m}{(x - r)^m}, \end{aligned}

and a factor $(x^2 + bx + c)^k$ contributes

\begin{aligned} &\frac{B_1 x + C_1}{x^2 + bx + c} + \cdots \\[0.4em] &\quad + \frac{B_k x + C_k}{(x^2 + bx + c)^k}. \end{aligned}

The identity holds at every $x$ with $Q(x) \ne 0$ . When $Q$ is a constant times a product of distinct linear factors $x - r_1, \dots, x - r_n$ , the decomposition is a sum of $n$ terms $A_i/(x - r_i)$ .

In plain words The decomposition undoes the addition of fractions. Each factor of the denominator gets its own simple fractions, a repeated factor gets one fraction for every power up to its multiplicity, and an irreducible quadratic gets a linear numerator. By linearity, each simple fraction can then be integrated on its own.

Example The integrand of the integral from $2$ to $3$ is proper, and $x^2 - 1 = (x - 1)(x + 1)$ has two distinct linear factors. Its zeros $1$ and $-1$ lie outside $[2, 3]$ , so the integrand is continuous on the interval. Write

\begin{aligned} &\frac{1}{(x - 1)(x + 1)} \\[0.4em] &\quad = \frac{A}{x - 1} + \frac{B}{x + 1}. \end{aligned}

Multiplying by $(x - 1)(x + 1)$ gives $1 = A(x + 1) + B(x - 1) = (A + B)x + (A - B)$ , so $A + B = 0$ and $A - B = 1$ , which gives $A = 1/2$ and $B = -1/2$ . Hence

\begin{aligned} \frac{1}{x^2 - 1} &= \frac{1}{2(x - 1)} \\[0.4em] &\quad - \frac{1}{2(x + 1)}, \end{aligned}

and at $x = 2$ the right side is $\tfrac12 - \tfrac16 = \tfrac13$ , which matches $1/(2^2 - 1)$ . On $[2, 3]$ both $x - 1$ and $x + 1$ are positive, so the logarithms need no absolute values, and

\begin{aligned} &\int_2^3 \frac{dx}{x^2 - 1} \\[0.4em] &\quad = \frac12\Bigl[\ln\frac{x - 1}{x + 1}\Bigr]_2^3 \\[0.4em] &\quad = \frac12\Bigl(\ln\frac12 - \ln\frac13\Bigr) \\[0.4em] &\quad = \frac12\ln\frac32, \end{aligned}

which is about $0.2027$ . As a check, the integrand decreases from $1/3$ at $x = 2$ to $1/8$ at $x = 3$ , so the integral over this interval of length $1$ lies between $0.125$ and $0.3333$ .

Comparing coefficients

Definition Two polynomials that take equal values at infinitely many real $x$ have the same coefficient for every power of $x$ , because their difference has infinitely many roots and must be the zero polynomial. After a proposed decomposition of $P/Q$ is multiplied by $Q$ , both sides are polynomials that agree wherever $Q(x) \ne 0$ , which is all but finitely many $x$ . So their coefficients agree, and equating the coefficients of each power of $x$ gives a linear system for the unknown constants. Because the cleared identity then holds for every real $x$ , it may also be evaluated at a zero of $Q$ , which often isolates one constant at a time. The original rational function is still undefined at that zero.

In plain words Clear the denominators, then match the $x^2$ terms, the $x$ terms and the constants on the two sides. Plugging in a zero of the denominator is a shortcut that works on the cleared polynomial identity, not on the original fraction.

Example To split $(x + 5)/\bigl((x - 1)(x + 2)\bigr)$ , write it as $A/(x - 1) + B/(x + 2)$ and clear the denominators,

\begin{aligned} &x + 5 \\[0.4em] &= A(x + 2) + B(x - 1) \\[0.4em] &= (A + B)x + (2A - B). \end{aligned}

Matching the coefficients gives $A + B = 1$ and $2A - B = 5$ , so $A = 2$ and $B = -1$ . The shortcut agrees. At $x = 1$ the identity reads $6 = 3A$ , and at $x = -2$ it reads $3 = -3B$ . Hence

\begin{aligned} &\frac{x + 5}{(x - 1)(x + 2)} \\[0.4em] &\quad = \frac{2}{x - 1} - \frac{1}{x + 2}, \end{aligned}

and at $x = 0$ both sides equal $-5/2$ . In the integral from $2$ to $3$ , the same shortcut applied to $1 = A(x + 1) + B(x - 1)$ gives $1 = 2A$ at $x = 1$ and $1 = -2B$ at $x = -1$ , the constants found by matching coefficients, although $1/(x^2 - 1)$ itself has no value at either point.

Logarithmic antiderivative

Definition For a constant $r$ , the function $\ln\lvert x - r \rvert$ is differentiable at every $x \ne r$ , with derivative $1/(x - r)$ . So on each interval that does not contain $r$ ,

\int \frac{dx}{x - r} = \ln\lvert x - r \rvert + C,

and more generally, for constants $p \ne 0$ and $q$ , an antiderivative of $1/(px + q)$ on each interval where $px + q \ne 0$ is $\tfrac1p\ln\lvert px + q \rvert$ . The constant of integration may differ on the two sides of the zero. On an interval where the argument is positive the absolute value may be dropped, and for positive $u$ and $v$ the identity $\ln u - \ln v = \ln(u/v)$ holds.

In plain words Each simple fraction $A/(x - r)$ integrates to $A$ times a logarithm. The absolute value makes the formula correct on either side of $r$ , and the logarithm of a quotient combines the two logarithms of a decomposition into one.

Example On $[2, 3]$ an antiderivative of $1/(x^2 - 1)$ is

\begin{aligned} F(x) &= \tfrac12\ln\lvert x - 1 \rvert \\[0.4em] &\quad - \tfrac12\ln\lvert x + 1 \rvert \\[0.4em] &= \tfrac12\ln\frac{x - 1}{x + 1}. \end{aligned}

Its values $F(3) = \tfrac12\ln\tfrac12 \approx -0.3466$ and $F(2) = \tfrac12\ln\tfrac13 \approx -0.5493$ are both negative, yet $F(3) - F(2) = \tfrac12\ln\tfrac32$ is positive, so the sign of an antiderivative says nothing about the sign of the integral. The arguments are divided, $\ln\tfrac12 - \ln\tfrac13 = \ln\tfrac32$ , never subtracted inside one logarithm, which would give $\ln\tfrac16$ . Differentiating confirms the coefficients, since $F'(x) = \tfrac{1}{2(x - 1)} - \tfrac{1}{2(x + 1)} = \tfrac{1}{x^2 - 1}$ for $x > 1$ . A linear factor with another slope brings its own coefficient, so the antiderivatives of $1/(2x + 1)$ on each side of $-1/2$ are $\tfrac12\ln\lvert 2x + 1 \rvert + C$ .

Poles outside the interval

Definition Let $P/Q$ be a rational function in lowest terms, so that $P$ and $Q$ have no common factor of positive degree. Its real poles are the real zeros of $Q$ . If no zero of $Q$ lies in $[a, b]$ , then $P/Q$ is continuous on $[a, b]$ , and for every antiderivative $F$ of $P/Q$ on an open interval containing $[a, b]$ ,

\begin{aligned} &\int_a^b \frac{P(x)}{Q(x)}\,dx \\[0.4em] &\quad = F(b) - F(a). \end{aligned}

If a pole $r$ lies in $[a, b]$ , inside or at an endpoint, the integral is improper and diverges. Near $r$ the function has the form $h(x)/(x - r)^m$ with $m \ge 1$ , $h$ continuous and $h(r) \ne 0$ , and comparison with the integral of $1/\lvert x - r \rvert^m$ , which diverges over every interval with endpoint $r$ , shows that the integral of $P/Q$ diverges. In that case $F(b) - F(a)$ is not the value of the integral.

In plain words Factoring shows where the denominator vanishes, but it does not remove those points from the domain. If they stay outside the interval, the integral is an ordinary one and endpoint subtraction is valid. If one lies in the interval, the region next to it has infinite area, and the logarithm formula hides that.

Example For $1/(x^2 - 1)$ the poles are $-1$ and $1$ , both outside $[2, 3]$ , which is why the integral from $2$ to $3$ can be evaluated by subtraction. On $[0, 3]$ the pole $1$ lies inside. Subtracting anyway gives $F(3) - F(0) = \tfrac12\ln\tfrac12 - 0 \approx -0.3466$ , a number attached to an integral that diverges. For $1 < x \le 3$ the factor $x + 1$ is at most $4$ , so

\frac{1}{x^2 - 1} \ge \frac{1}{4(x - 1)},

and the integral of $1/(x - 1)$ from $1$ to $3$ diverges. The integral of $1/x$ across $0$ behaves the same way. The difference $\ln\lvert 1 \rvert - \ln\lvert -1 \rvert = 0$ is not the value of $\int_{-1}^{1} dx/x$ , which diverges.

Check yourself

Evaluate the integral below with partial fractions.

\int_3^5 \frac{dx}{x^2 - 4}

Show the answer

The denominator factors as $(x - 2)(x + 2)$ , and its zeros $2$ and $-2$ lie outside $[3, 5]$ . From $1 = A(x + 2) + B(x - 2)$ , the value $x = 2$ gives $A = 1/4$ and the value $x = -2$ gives $B = -1/4$ . Both factors are positive on $[3, 5]$ , so

\begin{aligned} &\int_3^5 \frac{dx}{x^2 - 4} \\[0.4em] &\quad = \frac14\Bigl[\ln\frac{x - 2}{x + 2}\Bigr]_3^5 \\[0.4em] &\quad = \frac14\Bigl(\ln\frac37 - \ln\frac15\Bigr) \\[0.4em] &\quad = \frac14\ln\frac{15}{7} \approx 0.1905. \end{aligned}

The integrand decreases from $1/5$ to $1/21$ over an interval of length $2$ , so the value must lie between $2/21 \approx 0.0952$ and $0.4$ , and it does.

Evaluate the integral below. Why does a decomposition of the form $A/x + C/(x + 1)$ not work?

\int_1^2 \frac{dx}{x^2(x + 1)}

Show the answer

The factor $x$ appears squared, so the decomposition needs a term for each power,

\begin{aligned} &\frac{1}{x^2(x + 1)} \\[0.4em] &\quad = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1}. \end{aligned}

Clearing denominators gives $1 = Ax(x + 1) + B(x + 1) + Cx^2$ . The value $x = 0$ gives $B = 1$ , the value $x = -1$ gives $C = 1$ , and the coefficient of $x^2$ gives $A + C = 0$ , so $A = -1$ . The poles $0$ and $-1$ lie outside $[1, 2]$ , where $x$ and $x + 1$ are positive, so an antiderivative there is $-\ln x - 1/x + \ln(x + 1)$ , and

\begin{aligned} &\int_1^2 \frac{dx}{x^2(x + 1)} \\[0.4em] &\quad = \Bigl[\ln\frac{x + 1}{x} - \frac1x\Bigr]_1^2 \\[0.4em] &\quad = \frac12 + \ln\frac34 \approx 0.2123. \end{aligned}

The integrand decreases from $1/2$ to $1/12$ on an interval of length $1$ , and the value lies between those numbers. Without the term $B/x^2$ , clearing the denominator $x^2(x + 1)$ would give $1 = x\bigl(A(x + 1) + Cx\bigr)$ . This is a polynomial identity, so it would have to hold at $x = 0$ as well, where it reads $1 = 0$ . No constants $A$ and $C$ can work.

Someone evaluates the integral of $1/(x^2 - 1)$ from $0$ to $2$ as $F(2) - F(0) = \tfrac12\ln\tfrac13 \approx -0.5493$ , with the antiderivative $F$ from above. Is this the value of the integral?

Show the answer

No. The pole $x = 1$ lies inside $[0, 2]$ , so the integral is improper and has to be split at $1$ . On the right part

\begin{aligned} &\int_1^2 \frac{dx}{x^2 - 1} \\[0.4em] &\quad = \lim_{t \to 1^+} \int_t^2 \frac{dx}{x^2 - 1} \\[0.4em] &\quad = \lim_{t \to 1^+} \frac12\ln\frac{t + 1}{3(t - 1)}, \end{aligned}

and this limit is $+\infty$ , because the quotient $(t + 1)/\bigl(3(t - 1)\bigr)$ grows without bound as $t$ decreases to $1$ . So the integral from $0$ to $2$ diverges, and $-0.5493$ is not its value. The function $F$ is an antiderivative on $(-1, 1)$ and on $(1, \infty)$ separately, never on an interval that contains $1$ .

Partial fractions for a factorable denominator

Key terms

Rational function

Partial fraction decomposition

Comparing coefficients

Logarithmic antiderivative

Poles outside the interval

Check yourself

How this lesson connects

Builds on