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A proper fraction whose denominator factors into distinct linear factors splits into pieces that each integrate to a logarithm — and the degree check that has to come before the split.
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 can split a proper rational function with distinct linear factors into partial fractions, clearing the denominators and substituting each root in turn to read off its coefficient, and integrate the pieces to logarithms. You can also check the numerator's degree before you start, say what has to happen when it is too large, and use the sum of the coefficients as a free check on both of them.
You can factor a quadratic, and you know that $\int \dfrac{dx}{x - r}$ is $\ln|x - r|$. Partial fractions is the technique that turns a quotient you cannot integrate into a sum of quotients you can — and every piece it produces is one of those logarithms.
Proper fraction: one whose numerator has lower degree than its denominator. Only a proper fraction can be split.
Partial fractions: the pieces $\dfrac{A}{x - a} + \dfrac{B}{x + b}$ that add back to the original.
Clearing the denominators: multiplying both sides by the whole denominator to get a polynomial identity.
The cover-up move: substituting the root of one factor to make its partner's term vanish.
$\dfrac{1}{(x-2)(x+3)}$ has no antiderivative you can see. Each of $\dfrac{1}{x-2}$ and $\dfrac{1}{x+3}$ has one you have known for a year. So the plan is to write the first as a combination of the second two.
Write the split with unknowns and clear the denominators. Setting $$\frac{1}{(x-a)(x+b)} = \frac{A}{x-a} + \frac{B}{x+b}$$ and multiplying through gives $1 = A(x+b) + B(x-a)$. That is an identity — true for every $x$ — which is what makes the next step legal.
Substitute the roots, one at a time. Putting $x = a$ makes the $B$ term vanish and hands you $A$ immediately; putting $x = -b$ hands you $B$. Comparing coefficients also works and takes longer; substituting the roots is the move worth having as a reflex.
Then integrate. Each piece is a constant over a linear factor, so each gives a logarithm, and the answer is $A\ln|x-a| + B\ln|x+b| + C$.
The check that comes first. None of this is valid unless the numerator's degree is strictly below the denominator's. If it is not, do the division: $\dfrac{x^{2}}{x^{2}-1} = 1 + \dfrac{1}{x^{2}-1}$, and it is the remainder that gets split. Skipping that step leads to equations for $A$ and $B$ that have no solution at all.
Another way: picture
Think of the fraction as a chord and the pieces as the notes in it. You cannot say anything useful about the chord as a single object, but once it is written as separate notes each one is something you can already play. Integration is the playing; the splitting is just reading the chord correctly.
Another way: steps
| The numerator is | The coefficients | A free check |
|---|---|---|
| a constant | equal and opposite | their sum is $0$ |
| linear | unrelated in general | the sum equals the numerator's leading coefficient |
| the same degree as the bottom | not defined yet | divide first |
The middle row is the general rule and the top row is its commonest special case. Far from the roots, $\dfrac{A}{x-a} + \dfrac{B}{x+b}$ behaves like $\dfrac{A+B}{x}$, and that has to match what the original fraction does out there — which is where both checks come from.
Splitting an improper fraction. If the top is not of lower degree, the identity has no solution and the two substitutions will contradict each other. Divide first, every time.
Substituting into the original instead of the cleared identity. The roots are exactly the points where the original fraction is undefined. The substitution is legal only after the denominators have been cleared away.
Expecting the pieces to look like the original. They do not, and they should not. $\dfrac{1}{x^{2}-1}$ is one fraction and its split is two; adding them back is the way to check, not to be surprised.
$\dfrac{1}{(x-2)(x+3)} = \dfrac{A}{x-2} + \dfrac{B}{x+3}$, so $1 = A(x+3) + B(x-2)$.
Clear first.
$x = 2$ gives $1 = 5A$, so $A = \tfrac15$; $x = -3$ gives $1 = -5B$, so $B = -\tfrac15$.
One root at a time.
The integral is $\tfrac15\ln|x-2| - \tfrac15\ln|x+3| + C$, and $A + B = 0$ confirms both coefficients at once.
$\dfrac{x^{2}}{x^{2}-1}$: top and bottom are both quadratic, so no split is possible yet.
Check the degrees.
Divide: $\dfrac{x^{2}}{x^{2}-1} = 1 + \dfrac{1}{x^{2}-1}$, and only the second piece is split.
The quotient integrates on its own.
$\dfrac{1}{(x-1)(x+1)}$ splits as $\tfrac12$ over $x-1$ minus $\tfrac12$ over $x+1$, so the integral is $x + \tfrac12\ln\left|\dfrac{x-1}{x+1}\right| + C$.
Clear the denominators: $1 = A(x+4) + B(x-1)$, an identity in $x$.
Legal because it holds everywhere.
Substitute $x = 1$: $1 = 5A$, so $A = \tfrac15$.
One root kills one term.
Substitute $x = -4$: $1 = -5B$, so $B = -\tfrac15$ — and the sum being zero says both are right.
Put the steps of evaluating $\int \dfrac{dx}{x^{2} - x - 20}$ into the order you do them.
Number the steps in order (write the number in the box):
Write $\dfrac{1}{(x - 4)(x + 5)}$ as $\dfrac{A}{x - 4} + \dfrac{B}{x + 5}$. Give $A$, then $B$.
A = p, B = q
For $\dfrac{1}{(x - 1)(x + 1)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1}$, fill in the table.
| Value | |
|---|---|
| $A$ | |
| $B$ | |
| $A + B$ |
Having split $\dfrac{1}{(x - 3)(x + 5)}$ into two simple fractions, what does integrating give?
Split $\dfrac{x + 2}{(x - 3)(x + 6)}$ into $\dfrac{A}{x - 3} + \dfrac{B}{x + 6}$. What is $A$?
Answer:
Select every fraction that can be split into partial fractions as it stands, with no work done first.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of evaluating $\int \dfrac{dx}{x^{2} - x - 2}$ into the order you do them.
Number the steps in order (write the number in the box):
You can split a rational function and integrate the pieces. Without looking: what has to be true of the numerator's degree, which value of $x$ isolates the first coefficient, and what does every piece integrate to?
9. Your turn: split $\dfrac{1}{(x-1)(x+4)}$, step 3