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Every proper rational function splits into pieces of two shapes, so every rational function has an elementary antiderivative.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to check whether a rational function is proper and divide it when it is not, factor a denominator completely over the reals, write the decomposition the factors force — one constant per power of each linear factor and a linear numerator over each irreducible quadratic — find the constants by the cover-up rule and by comparing coefficients, integrate each piece to a logarithm, a power or an arctangent, and say why the number of unknown constants must equal the degree of the denominator.
You have added algebraic fractions since school: $\dfrac{1}{x-1} - \dfrac{1}{x-2}$ becomes $\dfrac{-1}{(x-1)(x-2)}$ over a common denominator. The result is harder to integrate than either piece was — $\ln|x-1| - \ln|x-2|$ is immediate, and nothing about $\dfrac{-1}{(x-1)(x-2)}$ is.
Partial fractions is that addition run backwards. It is not a calculus technique at all; it is an algebraic fact about rational functions, and the calculus is the three lines at the end where each piece is integrated by a formula you already know.
A fraction is proper when the numerator's degree is strictly less than the denominator's. Only proper fractions can be decomposed; an improper one is divided first.
A quadratic is irreducible over the real numbers when its discriminant is negative, so it has no real root and cannot be factored further. It keeps a linear numerator $Bx + C$ rather than a constant.
The cover-up rule finds the coefficient over a simple linear factor in one line: cover that factor in the original fraction and evaluate everything else at its root.
Theorem. Every proper rational function $\dfrac{P(x)}{Q(x)}$ over the real numbers is a sum of terms of the forms $$\frac{A}{(x - r)^k} \qquad\text{and}\qquad \frac{Bx + C}{(x^2 + bx + c)^k},$$ where the factors are those of $Q$ and $x^2 + bx + c$ is irreducible. The decomposition exists and is unique.
That is a genuine theorem, and it has a corollary worth stating: every rational function has an elementary antiderivative. Each of the two shapes integrates to a logarithm, a power, an arctangent, or a combination — never to anything new. No other family of functions in this course has that property.
Writing the shape. Factor $Q$ completely over the reals. Then:
The number of unknown constants always equals the degree of $Q$.
Finding the constants. Multiply through by $Q$ and compare. For simple linear factors the cover-up rule is fastest. For repeated factors and quadratics, substitute convenient values and then compare coefficients of the powers of $x$.
Integrating the pieces. $\displaystyle\int\frac{dx}{x - r} = \ln|x - r|$; $\displaystyle\int\frac{dx}{(x-r)^k} = \frac{(x-r)^{1-k}}{1-k}$ for $k > 1$; and $\dfrac{Bx + C}{x^2 + bx + c}$ is split against the derivative of the denominator, giving a logarithm plus an arctangent after completing the square — which is why the previous lesson came first.
Another way: picture
$\dfrac{1}{(x-1)(x+1)}$ has a spike at $x = 1$ and another at $x = -1$. The decomposition $\tfrac12\left(\dfrac{1}{x-1} - \dfrac{1}{x+1}\right)$ says the function is exactly the sum of the two spikes, each of which is a shifted copy of $1/x$. The picture of the method is: one simple pole at a time, each handled by itself.
Another way: steps
It is worth seeing that the theorem really does close the problem, because each shape has a known antiderivative and there are only three shapes.
Simple linear. $\displaystyle\int\frac{A\,dx}{x - r} = A\ln|x - r| + C$. The absolute value is not decoration: the antiderivative is needed on both sides of the pole.
Repeated linear. $\displaystyle\int\frac{A\,dx}{(x-r)^k} = \frac{-A}{(k-1)(x-r)^{k-1}} + C$ for $k \ge 2$. No logarithm appears — which is why a repeated factor must contribute its lower powers too, or the logarithm the function actually has would be missing.
Irreducible quadratic. Split $\dfrac{Bx + C}{x^2 + bx + c}$ as a multiple of $\dfrac{2x + b}{x^2 + bx + c}$ plus a constant over the same denominator. The first gives $\ln(x^2 + bx + c)$ — no absolute value needed, since an irreducible quadratic never changes sign — and the second gives an arctangent once the square is completed.
So the antiderivative of any rational function is a polynomial, plus logarithms, plus negative powers, plus arctangents. That is the complete list.
Decomposing an improper fraction. It cannot be done. Every proper piece tends to zero at infinity and an improper fraction does not; divide first.
A repeated factor given only its top power. $\dfrac{1}{(x-2)^2}$ needs $\dfrac{A}{x-2} + \dfrac{B}{(x-2)^2}$ in general. Omitting the first term makes the system unsolvable — which at least announces itself — or, worse, solvable with a wrong answer when the numerator happens to cooperate.
A constant over an irreducible quadratic. $\dfrac{A}{x^2+4}$ cannot produce the $x$ in a numerator like $x + 3$. The numerator must be linear.
Factoring an irreducible quadratic anyway. $x^2 + 4 = (x-2i)(x+2i)$ is true and outside this method, which works over the reals.
Cover-up on a repeated or quadratic factor. It gives the coefficient of the highest power of a repeated factor correctly and says nothing about the others.
Dropping the absolute value in the logarithm. $\ln(x - r)$ is undefined for $x < r$, and a definite integral to the left of the pole then appears to be impossible.
Learners often treat the form of a decomposition as something to try: put constants over everything, see whether the equations come out, adjust if they do not. That works often enough to become a habit and fails in exactly the cases that matter.
The shape is determined, and the reason is a counting argument. A decomposition of $P/Q$ must be able to represent every proper fraction with denominator $Q$, and those form a space of dimension $\deg Q$. So the shape needs exactly $\deg Q$ free constants — which is what you get by giving each power of each linear factor one constant and each irreducible quadratic two. Any fewer and some proper fractions are unreachable; any more and the constants are not unique.
That is why $\dfrac{1}{(x-2)^2}$ needs a term over $(x-2)$ as well: the space of proper fractions over $(x-2)^2$ is two-dimensional, and $\dfrac{B}{(x-2)^2}$ alone spans one dimension of it. And it is why $\dfrac{A}{x^2+4}$ is not enough: that denominator is degree two and needs two constants, so the numerator must be linear.
Count the degree of the denominator, count your unknowns, and if they disagree the shape is wrong before any arithmetic is done.
$\dfrac{3x + 1}{(x-1)^2} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2}$. Multiply by $(x-1)^2$: $3x + 1 = A(x-1) + B$.
Both powers, or the system has no solution.
$x = 1$ gives $B = 4$. Comparing the coefficient of $x$ gives $A = 3$.
One substitution, one comparison.
$\displaystyle\int\frac{3x+1}{(x-1)^2}dx = 3\ln|x-1| - \frac{4}{x-1} + C$. The logarithm comes only from the first-power term, which is exactly why it had to be there.
$\dfrac{x + 4}{x(x^2 + 4)} = \dfrac{A}{x} + \dfrac{Bx + C}{x^2 + 4}$, so $x + 4 = A(x^2 + 4) + (Bx + C)x$.
A linear numerator over the quadratic.
$x = 0$ gives $A = 1$. Comparing $x^2$: $A + B = 0$, so $B = -1$. Comparing $x$: $C = 1$.
Substitute where it is easy, compare for the rest.
$\displaystyle\int = \ln|x| - \tfrac12\ln(x^2+4) + \tfrac12\arctan\tfrac x2 + C$: a logarithm from each piece and an arctangent from the constant left over the quadratic.
Three standard integrals, and nothing else.
Proper, and the denominator factors: $\dfrac{1}{(x-2)(x+2)} = \dfrac{A}{x-2} + \dfrac{B}{x+2}$.
Check the degrees, then factor.
Cover-up: $A = \dfrac{1}{4}$ at $x = 2$, and $B = -\dfrac{1}{4}$ at $x = -2$.
One line each.
$\dfrac14\left[\ln|x-2| - \ln|x+2|\right]_3^4 = \dfrac14\left(\ln\dfrac13 - \ln\dfrac15\right) = \dfrac14\ln\dfrac53$.
Write $\dfrac{1}{(x - 8)(x - 2)} = \dfrac{A}{x - 8} + \dfrac{B}{x - 2}$. What is $A$?
The coefficient A is answer.
A proper fraction has denominator $(x + 3)(x - 3)$. What kind of decomposition does it need?
Decompose $\dfrac{x + 6}{(x - 4)(x - 7)} = \dfrac{A}{x - 4} + \dfrac{B}{x - 7}$ and check your work.
| Value | |
|---|---|
| $A$, from covering $(x - 4)$ | |
| $B$, from covering $(x - 7)$ | |
| $A + B$, which must equal the numerator's leading coefficient | 1 |
Match each denominator to the shape its decomposition must have.
| $\dfrac{A}{x - 4} + \dfrac{B}{x - 5}$ | $\dfrac{A}{x - 4} + \dfrac{B}{(x - 4)^{2}}$ | $\dfrac{Bx + C}{x^{2} + 16}$ | $\dfrac{A}{x} + \dfrac{Bx + C}{x^{2} + 16}$ | |
|---|---|---|---|---|
| $(x - 4)(x - 5)$ | ||||
| $(x - 4)^{2}$ | ||||
| $x^{2} + 16$ | ||||
| $x(x^{2} + 16)$ |
Put the steps of integrating $\dfrac{x^{2}}{(x - 6)(x + 6)}$ by partial fractions into the order you do them.
Number the steps in order (write the number in the box):
Dividing, $\dfrac{x^{2} + 2}{x^{2} - 64} = 1 + \dfrac{R}{x^{2} - 64}$. Find $R$.
The remainder R is answer.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Write $\dfrac{1}{(x - 1)(x - 3)} = \dfrac{A}{x - 1} + \dfrac{B}{x - 3}$. What is $A$?
The coefficient A is answer.
You can decompose a rational function and integrate every piece, and say why the shape of the decomposition is forced rather than chosen. Say in your own words why every rational function has an elementary antiderivative. Next: choosing between all six techniques from the form of the integrand alone.
10. Your turn: $\displaystyle\int_3^{4}\frac{dx}{x^{2} - 4}$, step 3
The interval avoids both poles, so the integral is proper.