Integration by Partial Fraction Decomposition

Intro

Once again, the words say it all. We're going to integrate a fraction or rational
expression by decomposing it into a sum of simpler fractions.

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Example 1

Say we need to integrate this: .

We know that there are 2 fractions with denominators (x – 1) and (x + 1)
that add to give the fraction we need to integrate. What we're missing is
the numerators of those 2 fractions.

So, we set up this equivalence

and then we solve for the values of A and B.

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From this equivalence, we know that the numerator on the left must equal
the numerator on the right -- so, 2 = A(x + 1) + B(x – 1)
Now we solve for A and B and then integrate the 2 simple fractions.

There are 2 ways to find the values for A and B. We can make a system of
2 equations in 2 unknowns (A and B) by balancing the coefficients or we can use
the most efficient method -- called "magic numbers". Here are both ways to do it.

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Balancing of Coefficients

When we expand the statement A(x + 1) + B(x – 1) = 2, we get
(A + B) x + (A – B) = 0 x + 2 which means that A + B = 0 and A – B = 2.
The solution to the system is A = 1 and B = – 1.

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Magic Number Method

Here, we assume that the statement A(x + 1) + B(x – 1) = 2 is true for all values of x.
We substitute the "magic" values of x (numbers that make brackets = 0) into the
equation and then we solve for A and B.

Setting x = – 1, we get A( – 1 + 1) + B( – 1 – 1) = 2 so – 2B = 2 and B = – 1.
Similarly, setting x = 1, A(1 + 1) + B(1 – 1) = 2 simplifies to 2A = 2 so A = 1.

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Now our integration question is

The solution is ln | x – 1 | – ln | x + 1 | = ln [(x – 1) / (x + 1) ] + C

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Since not all denominators are as simple as these, we formulate 2 approaches for the decomposition of fractions:

1/ Linear Factors: for these, such as the ones above, of the form px + q, we put
a constant in the numerator as we did above.

2/ Irreducible Quadratic Factors: for denominator factors such as 4x2 + 1, which we
can't factor, we put a linear expression Ax + B in the numerator.

The rule is that the numerator's degree must be 1 less than the denominator's.

Rules for Partial Fraction Decomposition:

If the denominator factor is linear (px + q) k , we make k fractions
of the form for n = 1, 2....k.

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If the denominator is an irreducible quadratic of the form (ax2 + bx + c) k,
we make k fractions of the form for n = 1, 2, 3 .......... k.

Notice that we make a fraction for each power of the denominator factor.

The first thing to do, before decomposing anything, is to check that the
numerator's degree is less than or equal to the denominator's
. If not,
we divide the fraction and then decompose the "remainder" fraction.

Say our division led to this result:

we only have to decompose the last part of this integrand since we can integrate 3x – 5
with the power rule.

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Example 2

Integrate

The denominator factors are (x 2 + 4) and (2x – 1), and since x 2 + 4 is an irreducible quadratic, we will need Ax + B in the numerator.

So now we know that x 2 – x – 21 = (Ax + B)(2x – 1) + C(x 2 + 4).

We have 1 magic number, x = ½ so we'll substitute it into the last expression.

( ½)² – ½ – 21 = (17/4) C
which makes C = – 5.

Now we use balancing of coefficients since we're out of magic numbers.
When we expand and reorder (Ax + B)(2x – 1) + C(x 2 + 4), we get
(2A + C) x 2 + (2B – A) x + (4C – B) = x 2 x – 21.

We already know that C is – 5 so we can solve for A by balancing the coefficients of x 2 on the left and right. We set 2A + C = 1 which makes 2A – 5 = 1, so A = 3.

To find B, we set 2B – A = – 1 (balancing the coefficients of x), we get B = 1.

So, our integration question becomes

Notice how we broke up the first fraction with the linear numerator to simplify the integration.
Now we integrate the 1st and 3rd by substitution, and the 2nd is just arcTan (½x).

Notice we use K as the constant of integration since we don't want to confuse it
with the C we used in the decomposition of the fraction.

Now we do one where the denominator factors are exponentiated.

Example 3

Integrate

We will make 2 fractions with (x – 1) and (x – 1)² as denominators, since we must
create a fraction for every power of the denominator factor.

So,

When we calculate the numerator of the combined fraction on the right of
this equation, we get that A(x – 1) + B = 6x – 11.
Since we have a magic number, namely x = 1, we substitute it to get B = – 5
and now it's easy to see that A = 6.

So, our integration question becomes

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Now we do one where the denominator factors are exponentiated and there's no magic numbers.

Example 4: Integrate

We will make 2 fractions with (x² + 1) and (x² + 1)² as denominators, since we must
create a fraction for every power of the denominator factor.

So,

When we calculate the numerator of the combined fraction on the right of
this equation, we get that (Ax + B)(x² + 1) + (Cx + D) = 5x³ – 3x² + 7x – 3.
When we expand, collect and reorder the coefficients we get

5x³ – 3x² + 7x – 3 = Ax³ + Bx² + (A + C) x + (B + D)

Since A is the only coefficient multiplying , A must = 5.
Similarly, since B is the only coefficient multiplying x², B must = – 3.
Since A + C = 7 (the coefficient of x), C = 2 and since B + D = – 3, D = 0.

So, our integration question becomes

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Practice

Use partial fraction decomposition and any other techniques of integration to evaluate:

 1) 2) 3) 4) 5)

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6) Find the area under the curve between x = 0 and x = 2.

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7) The region bounded by the graphs of , y = 0, x = 2 and x = 3
is revolved about the y-axis. Find the volume of the resulting solid.

Solutions

 1) 4 ln |x + 1| - 5 ln |x - 2| + ln |x - 3| + K 2) 3) 4) 5) by long division the integrand is The last integral is simply du / u if we set u = x² - x so the answer is or if we treat ln |x² - x| as ln |x| + ln |x - 1| the answer is 6) .The negative happens because the values of f(x) are all negative on the interval 0 to 2. Therefore, we take the absolute value of the result.The area under this curve = ½ ln 3.

7) Using shell method, the volume we want is

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