Split a rational function into simple pieces โ each one integrates to a logarithm, a power or an arctangent.
Adding fractions is something you have done since school: $\dfrac{4}{x-3}-\dfrac{1}{x+2}$ combines over a common denominator into $\dfrac{3x+11}{x^2-x-6}$. Integration asks us to run that process backwards. Faced with $\displaystyle\int\frac{3x+11}{x^2-x-6}\,dx$, no substitution and no integration by parts helps much โ but if we can recover the two simple fractions we started from, each one integrates in a single line to a natural logarithm.
The method of partial fractions does exactly this. It works for every rational function (a polynomial divided by a polynomial), and it is the reason that every rational function has an antiderivative built only from polynomials, logarithms and inverse tangents. Rational integrals turn up whenever a rate depends on a ratio โ population growth with a carrying capacity, chemical reactions between two reactants, the response of an electrical circuit โ so this is a technique engineers and scientists use constantly.
Proper and Improper Rational Functions
Partial fractions only work when the numerator is "smaller" than the denominator, in the sense of degree. So the very first thing to check is the degrees.
๐ Definition 5.1 โ Proper and improper rational functions
A rational function is a quotient $\dfrac{P(x)}{Q(x)}$ of polynomials with $Q$ not the zero polynomial. It is proper if $\deg P\lt\deg Q$, and improper if $\deg P\ge\deg Q$.
Every improper rational function can be rewritten by polynomial long division as
$$\frac{P(x)}{Q(x)}=S(x)+\frac{R(x)}{Q(x)},\qquad \deg R\lt\deg Q,$$
where the quotient $S$ is a polynomial (easy to integrate) and $\dfrac{R}{Q}$ is proper.
For example $\dfrac{x^2}{x+1}$ is improper (degree $2\ge1$), and division gives $\dfrac{x^2}{x+1}=x-1+\dfrac{1}{x+1}$; whereas $\dfrac{x+5}{x^2-1}$ is already proper.
Let $P$ and $Q$ be polynomials with real coefficients and $\deg P\lt\deg Q$. Factor $Q$ completely over the real numbers into linear factors $(ax+b)$ and irreducible quadratic factors $(ax^2+bx+c)$ with $b^2-4ac\lt0$ (factors may be repeated). Then $\dfrac{P}{Q}$ can be written, in exactly one way, as a sum of the following terms:
for each factor $(ax+b)^m$ in $Q$: $\quad\dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+\cdots+\dfrac{A_m}{(ax+b)^m}$;
for each factor $(ax^2+bx+c)^m$ in $Q$: $\quad\dfrac{B_1x+C_1}{ax^2+bx+c}+\cdots+\dfrac{B_mx+C_m}{(ax^2+bx+c)^m}$,
where the $A_i$, $B_i$, $C_i$ are real constants. Note that the number of unknown constants always equals $\deg Q$.
Check the degrees. If $\deg P\ge\deg Q$, do long division first.
Factor the denominator completely into linear and irreducible quadratic factors.
Write the form of the decomposition using Theorem 5.6 โ one unknown constant per degree of $Q$.
Find the constants โ by the cover-up method (substituting roots) and/or by equating coefficients.
Integrate term by term using the building blocks below, then check by recombining or differentiating.
๐ Theorem 5.7 โ The building-block integrals
For constants $a$, $k\ne0$ and $c\gt0$, on any interval where the integrand is continuous:
$$\int\frac{dx}{x-a}=\ln|x-a|+C,\qquad \int\frac{dx}{(x-a)^n}=\frac{-1}{(n-1)(x-a)^{n-1}}+C\ \ (n\ge2),$$
$$\int\frac{dx}{kx+b}=\frac1k\ln|kx+b|+C,\qquad \int\frac{x\,dx}{x^2+c^2}=\frac12\ln\left(x^2+c^2\right)+C,\qquad \int\frac{dx}{x^2+c^2}=\frac1c\tan^{-1}\frac{x}{c}+C.$$
Distinct Linear Factors
When $Q$ is a product of different linear factors $(x-r_1)(x-r_2)\cdots(x-r_n)$, the decomposition is $\dfrac{A_1}{x-r_1}+\cdots+\dfrac{A_n}{x-r_n}$, and every term integrates to a logarithm. There are two ways to find the constants.
๐งญ Two ways to find the constants
Equating coefficients. Multiply both sides by $Q(x)$ to clear all fractions, expand, and match the coefficients of $1,x,x^2,\dots$ on both sides. This gives a linear system for the constants. It always works.
The cover-up method (for a non-repeated linear factor $x-r$). Since $\dfrac{P(x)}{Q(x)}=\dfrac{A}{x-r}+(\text{terms that stay finite at }r)$, multiplying by $(x-r)$ and letting $x\to r$ gives
$$A=\lim_{x\to r}(x-r)\frac{P(x)}{Q(x)}.$$
In practice: cover up the factor $(x-r)$ in the denominator with your finger, and substitute $x=r$ into what is left. Interactive 1 shows this limit happening.
Check & factor Numerator degree $1\lt2$, so the function is proper. $x^2-x-6=(x-3)(x+2)$, two distinct linear factors, so $$\frac{3x+11}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}.$$
Equate coefficients Multiply by $(x-3)(x+2)$: $3x+11=A(x+2)+B(x-3)=(A+B)x+(2A-3B)$. Matching: $A+B=3$ and $2A-3B=11$. From the first, $B=3-A$; then $2A-9+3A=11$, so $A=4$, $B=-1$.
Or cover up Cover $(x-3)$ and put $x=3$: $A=\dfrac{3(3)+11}{3+2}=\dfrac{20}{5}=4$. Cover $(x+2)$ and put $x=-2$: $B=\dfrac{3(-2)+11}{-2-3}=\dfrac{5}{-5}=-1$. Same answer, much faster.
Interactive 1 โ Coefficient Finder: the Cover-up Is a Limit
Drag the $A$ and $B$ sliders until the dashed sum $\dfrac{A}{x-a}+\dfrac{B}{x-b}$ lies exactly on the black curve $f$. Or press Play: the purple curve is $(x-a)f(x)$ โ the cover-up โ and as the dot slides to $x=a$ its height becomes $A$; then the green curve $(x-b)f(x)$ delivers $B$.
Divide Degree $3\ge2$: improper. Long division: $x^3-4x+1=(x^2-3x+2)(x+3)+(3x-5)$. (Check: $(x^2-3x+2)(x+3)=x^3-7x+6$, and $x^3-7x+6+3x-5=x^3-4x+1$ โ.) So $$\frac{x^3-4x+1}{x^2-3x+2}=x+3+\frac{3x-5}{(x-1)(x-2)}.$$
Why divide? Had we skipped the division and written $\dfrac{A}{x-1}+\dfrac{B}{x-2}$ for the original function, the right-hand side would tend to $0$ as $x\to\infty$ while the left-hand side grows like $x$ โ no constants $A$, $B$ could ever work.
Repeated and Quadratic Factors
Repeated linear factors
If a linear factor appears to a power, say $(x-r)^m$, then Theorem 5.6 demands one term for every power from $1$ up to $m$: $\dfrac{A_1}{x-r}+\dfrac{A_2}{(x-r)^2}+\cdots+\dfrac{A_m}{(x-r)^m}$. Cover-up still gives the coefficient of the highest power directly; the others come from equating coefficients (or substituting convenient values of $x$).
Form Proper ($2\lt3$). $x^3-2x^2=x^2(x-2)$: the factor $x$ is repeated twice. $$\frac{5x^2-3x+2}{x^2(x-2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-2}.$$
Cover-up For $C$: cover $(x-2)$, put $x=2$: $C=\dfrac{20-6+2}{4}=4$. For $B$ (the highest power of $x$): cover $x^2$, put $x=0$: $B=\dfrac{2}{0-2}=-1$.
Equate Clearing denominators: $5x^2-3x+2=Ax(x-2)+B(x-2)+Cx^2$. The $x^2$-coefficients give $5=A+C$, so $A=1$. (Check the $x$-coefficient: $-3=-2A+B=-2-1$ โ.)
Integrate $$\int\left(\frac1x-\frac{1}{x^2}+\frac{4}{x-2}\right)dx=\ln|x|+\frac1x+4\ln|x-2|+C.$$ Notice that the repeated factor produced a power term $\frac1x$, not a logarithm.
โ ๏ธ Common mistakes
1. Forgetting to divide. If $\deg P\ge\deg Q$, the decomposition does not exist until you divide (Example 2).
2. Too few terms for a repeated factor. $\dfrac{1}{x^2(x+1)}\ne\dfrac{A}{x^2}+\dfrac{B}{x+1}$ for any $A,B$: three unknowns are needed because $\deg Q=3$. Count your constants!
3. A constant over a quadratic. Over an irreducible quadratic the numerator must be linear, $Bx+C$ โ not just a constant.
4. Cover-up on a repeated factor. Covering $x^2$ in Example 3 gives the coefficient of $\frac{1}{x^2}$, not of $\frac1x$.
5. Dropping absolute values. $\int\frac{dx}{x-3}=\ln|x-3|+C$; the bars matter for $x\lt3$.
Irreducible quadratic factors
๐ Definition 5.2 โ Irreducible quadratic
A quadratic $ax^2+bx+c$ with real coefficients is irreducible if it cannot be factored into real linear factors โ equivalently, its discriminant is negative, $b^2-4ac\lt0$, so it has no real roots. Examples: $x^2+4$, $x^2+2x+5=(x+1)^2+4$.
An irreducible quadratic contributes a term $\dfrac{Bx+C}{ax^2+bx+c}$. To integrate it, complete the square and split the numerator into a multiple of the derivative of the denominator (giving a logarithm) plus a constant (giving an arctangent).
โ๏ธ Example 4 โ One linear and one quadratic factor
Irreducible? $b^2-4ac=4-20=-16\lt0$, so this is already a single partial fraction. Complete the square: $x^2+2x+5=(x+1)^2+4$.
Split the numerator The derivative of the denominator is $2x+2$. Write $x+3=\tfrac12(2x+2)+2$: $$\int\frac{x+3}{x^2+2x+5}\,dx=\frac12\int\frac{2x+2}{x^2+2x+5}\,dx+2\int\frac{dx}{(x+1)^2+4}.$$
Integrate The first integral is $\ln\left(x^2+2x+5\right)$ (the denominator is always positive, so no bars are needed). In the second put $u=x+1$: $\displaystyle2\int\frac{du}{u^2+4}=2\cdot\frac12\tan^{-1}\frac u2$. Hence $$\int\frac{x+3}{x^2+2x+5}\,dx=\frac12\ln\left(x^2+2x+5\right)+\tan^{-1}\frac{x+1}{2}+C.$$
Interactive 2 โ Decomposition Visualiser
The black curve is the rational function; the coloured curves are its partial fractions. A vertical probe sweeps across: at every $x$ the coloured values add up to the black value โ even next to the vertical asymptotes, where one term explodes and carries $f$ with it. Each coloured curve is something you can integrate on sight.
$x$โ
value of each termโ
sum of the termsโ
$f(x)$โ
An Application: Growth with a Ceiling
A population $P(t)$ that grows at relative rate $r$ but is limited by a carrying capacity $K$ obeys the logistic model $\dfrac{dP}{dt}=rP\left(1-\dfrac PK\right)$ (studied fully in Chapter 7). Turning it around, $\dfrac{dt}{dP}=\dfrac{1}{rP(1-P/K)}$, so the time needed to grow from $P_0$ to $P_1$ is an integral of a rational function โ a job for partial fractions.
โ๏ธ Example 6 โ How long until the fish stock recovers?
A protected lake can support $K=1000$ fish, and the stock grows logistically with $r=0.4$ per year. How long does it take the stock to grow from $100$ to $500$ fish? The time is
$$T=\int_{100}^{500}\frac{dP}{0.4\,P\left(1-\frac{P}{1000}\right)}.$$
๐ก Show solutionโถ
Simplify $P\left(1-\frac{P}{1000}\right)=\dfrac{P(1000-P)}{1000}$, so $T=\dfrac{1}{0.4}\displaystyle\int_{100}^{500}\frac{1000}{P(1000-P)}\,dP$.
Integrate $\displaystyle\int\left(\frac1P+\frac{1}{1000-P}\right)dP=\ln P-\ln(1000-P)=\ln\frac{P}{1000-P}$ (both $P$ and $1000-P$ are positive here). Therefore $$T=2.5\left[\ln\frac{P}{1000-P}\right]_{100}^{500}=2.5\left(\ln1-\ln\frac19\right)=2.5\ln9\approx5.49\text{ years}.$$
Interpret About five and a half years. Check the sign: $\ln9\gt0$, so the time is positive, as it must be. Exercise IV.7 asks the same question for a different stock.
Exercises
I
Distinct linear factors. Decompose into partial fractions and integrate.
8 Show that, for a constant $a\gt0$, $\displaystyle\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C$.
๐ก Show solutionsโถ
1 $\dfrac{1}{(x-1)(x+1)}=\dfrac{1/2}{x-1}-\dfrac{1/2}{x+1}$ (cover-up at $x=1$ and $x=-1$). $$\int\frac{dx}{x^2-1}=\frac12\ln|x-1|-\frac12\ln|x+1|+C=\frac12\ln\left|\frac{x-1}{x+1}\right|+C.$$
5 $2x^2+x-1=(2x-1)(x+1)$. Cover $(2x-1)$, put $x=\tfrac12$: $A=\dfrac{4.5}{1.5}=3$; cover $(x+1)$, put $x=-1$: $B=\dfrac{3}{-3}=-1$. So $\dfrac{3}{2x-1}-\dfrac{1}{x+1}$, and since $\int\frac{dx}{2x-1}=\frac12\ln|2x-1|$, $$\int\frac{x+4}{2x^2+x-1}\,dx=\frac32\ln|2x-1|-\ln|x+1|+C.$$
8 $\dfrac{1}{(x-a)(x+a)}=\dfrac{A}{x-a}+\dfrac{B}{x+a}$ with $A=\dfrac{1}{2a}$, $B=\dfrac{1}{-2a}$. Integrating, $\dfrac{1}{2a}\left(\ln|x-a|-\ln|x+a|\right)+C=\dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right|+C$. (Part 1 is the case $a=1$.)
II
Improper rational functions (divide first) and repeated linear factors. Evaluate.
2 $x^3+2=x(x^2-1)+(x+2)$, so the integrand is $x+\dfrac{x+2}{(x-1)(x+1)}$; cover-up: $\dfrac{3/2}{x-1}-\dfrac{1/2}{x+1}$. $$\int\frac{x^3+2}{x^2-1}\,dx=\frac{x^2}{2}+\frac32\ln|x-1|-\frac12\ln|x+1|+C.$$
8 $x^4=(x^2-1)(x^2+1)+1$, so the integrand is $x^2+1+\dfrac{1}{x^2-1}$. With Ex. I.1: $$\int\frac{x^4}{x^2-1}\,dx=\frac{x^3}{3}+x+\frac12\ln\left|\frac{x-1}{x+1}\right|+C.$$
4 $x^2+2x+10=(x+1)^2+9$ and $x+5=\tfrac12(2x+2)+4$: $$\int\frac{x+5}{x^2+2x+10}\,dx=\frac12\ln\left(x^2+2x+10\right)+\frac43\tan^{-1}\frac{x+1}{3}+C.$$
7 $x^4-1=(x-1)(x+1)(x^2+1)$. Cover-up: at $x=1$: $\dfrac{4}{2\cdot2}=1$; at $x=-1$: $\dfrac{4}{(-2)(2)}=-1$. Then $\dfrac{1}{x-1}-\dfrac1{x+1}=\dfrac{2}{x^2-1}$, and $\dfrac{4}{x^4-1}-\dfrac{2}{x^2-1}=\dfrac{4-2(x^2+1)}{x^4-1}=\dfrac{-2}{x^2+1}$. So $$\int\frac{4\,dx}{x^4-1}=\ln\left|\frac{x-1}{x+1}\right|-2\tan^{-1}x+C.$$
IV
Definite integrals, substitutions and applications.
1 $\displaystyle\int_2^4\frac{dx}{x^2-1}$
2 $\displaystyle\int_0^1\frac{x}{x^2+3x+2}\,dx$
3 $\displaystyle\int_0^1\frac{x^3}{x^2+1}\,dx$
4 Find the area of the region under $y=\dfrac{1}{x(x+1)}$ from $x=1$ to $x=3$.
5 Use the substitution $u=e^x$ to evaluate $\displaystyle\int\frac{e^x}{(e^x+1)(e^x+2)}\,dx$.
6Applied (chemistry). In the reaction $\mathrm{A}+\mathrm{B}\to\mathrm{C}$ the amount $x$ (mol/L) of product satisfies $\dfrac{dx}{dt}=k(5-x)(3-x)$ with $k=0.01$ L/(molยทmin) and $x(0)=0$. The time to reach $x=2$ is $t=\dfrac1k\displaystyle\int_0^2\frac{dx}{(5-x)(3-x)}$. Find it.
7Applied (ecology). A fish stock grows logistically with $r=0.5$ per year and carrying capacity $K=2000$. How long does it take to grow from $200$ to $1500$ fish?
8Conceptual. A student writes $\dfrac{x^2+1}{x(x-1)}=\dfrac Ax+\dfrac B{x-1}$. Explain what is wrong, and give the correct decomposition.
๐ก Show solutionsโถ
1 By Ex. I.1: $\dfrac12\left[\ln\dfrac{x-1}{x+1}\right]_2^4=\dfrac12\left(\ln\dfrac35-\ln\dfrac13\right)=\dfrac12\ln\dfrac95\approx0.2939$.
4 The function is positive on $[1,3]$. $\dfrac{1}{x(x+1)}=\dfrac1x-\dfrac1{x+1}$: area $=\Big[\ln\dfrac{x}{x+1}\Big]_1^3=\ln\dfrac34-\ln\dfrac12=\ln\dfrac32\approx0.4055$ square units.
5 $u=e^x$, $du=e^xdx$: $\displaystyle\int\frac{du}{(u+1)(u+2)}=\int\left(\frac1{u+1}-\frac1{u+2}\right)du=\ln\frac{u+1}{u+2}$. So the answer is $\ln\dfrac{e^x+1}{e^x+2}+C$ (no bars needed: both are positive).
6 $\dfrac{1}{(5-x)(3-x)}=\dfrac{1}{2}\left(\dfrac{1}{3-x}-\dfrac{1}{5-x}\right)$ (cover-up at $x=3$ and $x=5$). For $0\le x\le2$: $\displaystyle\int_0^2=\frac12\Big[\ln(5-x)-\ln(3-x)\Big]_0^2=\frac12\left(\ln3-0-\ln5+\ln3\right)=\frac12\ln\frac95$. So $t=100\cdot\tfrac12\ln\tfrac95=50\ln1.8\approx29.4$ minutes.
7 As in Example 6, $T=\dfrac1r\Big[\ln\dfrac{P}{K-P}\Big]_{200}^{1500}=2\left(\ln\dfrac{1500}{500}-\ln\dfrac{200}{1800}\right)=2(\ln3+\ln9)=2\ln27\approx6.59$ years.
8 The function is improper (degree $2$ over degree $2$), so it must be divided first: $\dfrac{x^2+1}{x(x-1)}=1+\dfrac{x+1}{x(x-1)}$. Cover-up on the remainder: $x=0$ gives $-1$, $x=1$ gives $2$. Correct decomposition: $1-\dfrac1x+\dfrac{2}{x-1}$.
๐ฌ ResearchโTeaching NexusResearch-tutored ยท optional, about 30 minutes
Research question. How long does a second-order chemical reaction take to reach a target conversion โ and how sensitive is that time to the rate constant?
Why it matters. Process engineers size batch reactors (for biodiesel, pharmaceuticals or water treatment) by the time a reaction needs. For $\mathrm{A}+\mathrm{B}\to\mathrm{C}$ the rate law $\dfrac{dx}{dt}=k(a-x)(b-x)$ leads straight to a partial-fraction integral.
What to do.
For $a\ne b$, separate the variables and use partial fractions to show $t(x)=\dfrac{1}{k(a-b)}\ln\dfrac{b(a-x)}{a(b-x)}$ for $0\le x\lt b$.
Sample case: $a=1.0$ mol/L, $b=0.8$ mol/L, $k=0.05$ L/(molยทs). Find the time to convert half of B ($x=0.4$).
Change $k$ by $\pm10\%$ and recompute. How does $t$ scale with $k$? Explain from the formula.
Check your value numerically: approximate $\int_0^{0.4}\frac{dx}{k(a-x)(b-x)}$ with a spreadsheet using 20 thin rectangles.
Tools: Calculator, spreadsheet (or a few lines of Python).
Hand in: A half-page note with the derivation, the times for the three values of $k$, the numerical check, and one sentence on what happens as $x\to b$.
Going further. What happens when $a=b$? (The decomposition changes โ the factor becomes repeated.) Find $t(x)$ in that case.
Instructor note: $t(0.4)=100\ln1.2\approx18.2$ s; $t$ is exactly proportional to $1/k$ (about $16.6$ s and $20.3$ s for $k=0.055$ and $0.045$). As $x\to b^-$ the time tends to infinity. For $a=b$: $t=\frac1k\left(\frac{1}{a-x}-\frac1a\right)$.
๐งช Self-check quiz
0/5 correct
Q1. Which is the correct form of the decomposition of $\dfrac{3x+1}{(x-1)^2\left(x^2+4\right)}$?
Q2. Before decomposing $\dfrac{x^3+1}{x^2-4}$ you must firstโฆ
Q3. If $\dfrac{x+5}{(x-1)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}$, then $A=$