How do I decompose 1/(x^2 - 1)?

Put 1 in the numerator box and x^2 - 1 in the denominator box. The denominator factors as (x - 1)(x + 1), so the answer is 1/(2(x - 1)) - 1/(2(x + 1)). The halves are exact: the page solves for them as whole-number pairs and writes the 2 into the denominator the way a textbook does, rather than printing a decimal that no longer adds back to 1.

What happens with a repeated factor like (x - 1) squared?

A factor that divides the denominator more than once gets one part at every power up to its multiplicity, which is the step most hand methods skip. Type x + 3 over (x - 1)^2 and you get 1/(x - 1) + 4/(x - 1)^2. If the numerator over a lower power solves to 0 that part is not printed, so 1 over (x - 1)^2 comes back as itself: it is already a partial fraction.

How does it handle an irreducible quadratic?

A quadratic with no rational root stays whole and takes a numerator of degree 1 rather than a constant. Type 3x + 5 over (x - 1)(x^2 + 1) and the answer is 4/(x - 1) - (4x + 1)/(x^2 + 1). The page never splits x^2 + 1 into complex or square-root factors, because an answer written in radicals is one you cannot check by multiplying it back.

What if the numerator has the higher degree?

Then the fraction is improper and it has to be divided before it can be split, which is where most wrong answers start. The page does that first and keeps the polynomial quotient beside the fractions instead of folding it into one of them: x^3 over x^2 - 1 comes back as x + 1/(2(x - 1)) + 1/(2(x + 1)), with the quotient x shown in its own cell.

Why does it say unsupported instead of giving an answer?

Because the denominator has a piece that is neither linear nor an irreducible quadratic, and those are the only shapes a partial fraction is made of. A denominator such as x^3 + x + 1 has no rational root and no rational quadratic factor, so it is returned unchanged with the piece that stopped the split named in words. Nothing is approximated, and no decimal factorisation is invented to fill the gap.

Does it cancel a common factor first?

Yes, and it tells you. Type x - 1 over x^2 - 1 and the page cancels x - 1, answers 1/(x + 1), and says the parts match your fraction wherever x - 1 is not 0, because your original fraction has no value at that point. Leaving the factor in would put a 0 on top of one of the parts, which is where the usual confusion comes from.

How large an expression will it take, and what can I type?

Both boxes take a power up to 8 and are read up to 500 characters. You can type an expression, a product such as (x - 1)^2(x + 2), or a list of coefficients separated by commas from the highest power down. Fractions such as 2/3 and decimals such as 0.75 are read as the exact rationals they are, and pasted maths works too: superscripts and a typographic minus sign are read as the powers and signs they stand for.

Does it show the work, like the cover-up method?

It shows the result and the evidence for it rather than one hand method, because the cover-up trick only reaches a distinct linear factor. You get the parts as a drawn fraction, the exact numerator of each one, the factor it sits over and its power, then the line that multiplies every part by what it was missing and adds them back to your numerator, with the difference printed as 0. That last line is the part you can check without trusting the page.

Partial Fraction Calculator

Type a numerator and a denominator and the page splits the fraction into its parts over the rationals. It cancels anything dividing both first, divides out the polynomial part when the top is not of lower degree than the bottom, factors the denominator, and solves for one numerator per factor and per power of that factor. Coefficients are held as exact fractions rather than decimals, so a half is a half all the way to the answer. The part most calculators get wrong is the end of the job: here the parts are added back over the common denominator and subtracted from the numerator you typed, and unless that difference is the zero polynomial nothing is shown at all. A denominator with a piece that is neither linear nor an irreducible quadratic is the one case a partial fraction cannot use, and it comes back unchanged and labelled unsupported instead of approximated.

rational expressiondegree 1 over degree 3

      3x + 5      
------------------
 (x - 1)(x^2 + 1) 

proper: numerator degree < denominator

decompositionexact

   4         4x + 1
-------  -  ---------
 x - 1       x^2 + 1

recombination
  4(x^2 + 1) - (4x + 1)(x - 1) = 3x + 5
  difference from your numerator   0

Numerator

Denominator

Terms: 2
Denominator: linear × irreducible quadratic
Quotient: None, proper
Check: Exact

x^2 + 1 has no rational root, so it stays whole and its part carries a numerator of degree 1 rather than a constant.

1/(x² - 1)two linear factors
1/(2(x - 1)) - 1/(2(x + 1))
(x + 3)/(x - 1)²a repeated linear factor
1/(x - 1) + 4/(x - 1)^2
x³/(x² - 1)improper, divided first
x + 1/(2(x - 1)) + 1/(2(x + 1))

Accuracy. Decomposition over rational linear and irreducible quadratic factors, with repeated factors kept at every multiplicity. An improper fraction is divided first and its polynomial quotient is shown separately rather than folded in. The page recombines the parts symbolically and checks the difference against your input is zero before showing them, and a denominator it cannot factor inside its degree bound is returned unchanged and labelled unsupported.

Every part, one by one

A linear factor takes a constant on top. An irreducible quadratic takes a numerator of degree one, because that is the most a numerator can be and still sit below it. A factor that divides the denominator more than once takes one part at every power up to its multiplicity, and the numerator column is the exact rational the page solved for, before any fraction was cleared into the denominator.

Part	Over	Power	Numerator	Factor
+ 4/(x - 1)	x - 1	1	4	linear, root 1
- (4x + 1)/(x^2 + 1)	x^2 + 1	1	-4x - 1	irreducible quadratic, discriminant -4

How this is worked out

Coefficients are held as whole-number pairs, a numerator over a denominator, so nothing is ever rounded and a fraction you type survives to the answer. Anything dividing both the top and the bottom comes out first. If the top is not of lower degree than the bottom, the fraction is improper and is divided: the polynomial part is shown on its own and only what is left goes over the denominator. The denominator is then factored over the rationals, every rational root by the rational root theorem and every quadratic by a search inside a bound on the size of a root. Both searches are finite, and one that runs out of steps says so rather than calling what is left irreducible.

One unknown polynomial is set up for each factor and each power of it, of degree below that factor, and the whole system is solved exactly by elimination over the rationals. Then the parts are put back over the common denominator, added up and subtracted from the numerator you typed. That difference has to be the zero polynomial or nothing is shown at all, which is why the check cell reads Exact rather than quoting a percentage.

A denominator whose pieces are not linear and not quadratic is the one case a partial fraction cannot use. There the page returns your expression unchanged and labels the result unsupported, naming the piece that stopped it, because a decomposition over factors nobody can write down in rationals would be a guess wearing the clothes of an answer.

What you can type

One letter for the variable, whole numbers, fractions such as 2/3, decimals such as 0.75, the signs + - * / ^ and brackets. Multiplication can be left out where it is obvious, so 2x and (x - 1)(x + 2) read the way they are written, and pasted maths works too: superscripts such as x² and a typographic minus are read as the powers and signs they stand for. Both boxes take a power up to 8 and are read up to 500characters. A list separated by commas and holding no letter, such as 1, 0, -1, is read as coefficients from the highest power down. Dividing inside a box only works by a number: the fraction itself belongs in the two boxes, one part in each.

Common questions

How do I decompose 1/(x^2 - 1)?: Put 1 in the numerator box and x^2 - 1 in the denominator box. The denominator factors as (x - 1)(x + 1), so the answer is 1/(2(x - 1)) - 1/(2(x + 1)). The halves are exact: the page solves for them as whole-number pairs and writes the 2 into the denominator the way a textbook does, rather than printing a decimal that no longer adds back to 1.
What happens with a repeated factor like (x - 1) squared?: A factor that divides the denominator more than once gets one part at every power up to its multiplicity, which is the step most hand methods skip. Type x + 3 over (x - 1)^2 and you get 1/(x - 1) + 4/(x - 1)^2. If the numerator over a lower power solves to 0 that part is not printed, so 1 over (x - 1)^2 comes back as itself: it is already a partial fraction.
How does it handle an irreducible quadratic?: A quadratic with no rational root stays whole and takes a numerator of degree 1 rather than a constant. Type 3x + 5 over (x - 1)(x^2 + 1) and the answer is 4/(x - 1) - (4x + 1)/(x^2 + 1). The page never splits x^2 + 1 into complex or square-root factors, because an answer written in radicals is one you cannot check by multiplying it back.
What if the numerator has the higher degree?: Then the fraction is improper and it has to be divided before it can be split, which is where most wrong answers start. The page does that first and keeps the polynomial quotient beside the fractions instead of folding it into one of them: x^3 over x^2 - 1 comes back as x + 1/(2(x - 1)) + 1/(2(x + 1)), with the quotient x shown in its own cell.
Why does it say unsupported instead of giving an answer?: Because the denominator has a piece that is neither linear nor an irreducible quadratic, and those are the only shapes a partial fraction is made of. A denominator such as x^3 + x + 1 has no rational root and no rational quadratic factor, so it is returned unchanged with the piece that stopped the split named in words. Nothing is approximated, and no decimal factorisation is invented to fill the gap.
Does it cancel a common factor first?: Yes, and it tells you. Type x - 1 over x^2 - 1 and the page cancels x - 1, answers 1/(x + 1), and says the parts match your fraction wherever x - 1 is not 0, because your original fraction has no value at that point. Leaving the factor in would put a 0 on top of one of the parts, which is where the usual confusion comes from.
How large an expression will it take, and what can I type?: Both boxes take a power up to 8 and are read up to 500 characters. You can type an expression, a product such as (x - 1)^2(x + 2), or a list of coefficients separated by commas from the highest power down. Fractions such as 2/3 and decimals such as 0.75 are read as the exact rationals they are, and pasted maths works too: superscripts and a typographic minus sign are read as the powers and signs they stand for.
Does it show the work, like the cover-up method?: It shows the result and the evidence for it rather than one hand method, because the cover-up trick only reaches a distinct linear factor. You get the parts as a drawn fraction, the exact numerator of each one, the factor it sits over and its power, then the line that multiplies every part by what it was missing and adds them back to your numerator, with the difference printed as 0. That last line is the part you can check without trusting the page.

Decomposition over rational linear and irreducible quadratic factors, with repeated factors kept at every multiplicity. An improper fraction is divided first and its polynomial quotient is shown separately rather than folded in. The page recombines the parts symbolically and checks the difference against your input is zero before showing them, and a denominator it cannot factor inside its degree bound is returned unchanged and labelled unsupported.