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Example: Compute ( (3x3–3x2+x+1)/(x4–2x3+2x2–2x+1) dx.

First step: Factor the denominator. Any obvious roots?

..?..

x=1 is a root, so we can divide by x–1:

(x4–2x3+2x2–2x+1)/(x–1) = x3–x2+x–1

Any obvious roots of x3–x2+x–1?

..?..

x=1 is a root (again), so we can divide by x–1 (again):

(x3–x2+x–1)/(x–1) = x2+1

Any obvious roots in x2+1? …

..?..

Not using real numbers!

So the denominator of the integrand factors as

(x–1)2(x2+1).

This means we can write p(x)/q(x) in this case as a sum of three simpler rational functions, with denominators...

..?..

x–1, (x–1)2, and x2+1, whose numerators are polynomials of low degree.

More specifically, there exist constants A, B, C, D such that

p(x)/q(x) = (3x3–3x2+x+1)/(x4–2x3+2x2–2x+1) equals

A/(x–1) + B/(x–1)2 + (Cx+D)/(x2+1).

(See Case II in Stewart for the handling of repeated linear factors, and Case III for the handling of non-repeated quadratic factors.)

To find the right A, B, C, and D, multiply through by q(x):

(*) 3x3–3x2+x+1 = A(x–1)(x2+1) + B(x2+1) + (Cx+D)(x–1)2

From there, there are a couple of ways to go:

Method: Expand (*), equate coefficients, and solve.

3x3–3x2+x+1 = A(x3–x2+x–1) + B(x2+1) + (Cx+D)(x2–2x+1)

= (A+C)x3 + (–A+B–2C+D)x2 + (A+C–2D)x + (–A+B+D)

So we need to find A, B, C, D satisfying

(1) A+C = 3

(2) –A+B–2C+D = –3

(3) A+C–2D = 1

(4) –A+B+D = 1

Yuck! (See the examples in the book, which are doable by this method.)

As way of getting more useful linear equations, I suggest:

Trick: Substitute the roots of q(x) into the equation (*) and its derivatives.

Plugging x=1 into

(*) 3x3–3x2+x+1 = A(x–1)(x2+1) + B(x2+1) + (Cx+D)(x–1)2

we get

2 = A(0) + B(2) + (C+D)(0)

so B = 1.

The derivative of (*) is

9x2–6x+1 = A(3x2–2x+1) + B(2x) + C(x–1)2+ (Cx+D)2(x–1)

and plugging x=1 into this we get

4 = A(2) + B(2) + C(0) + D(0)

so 4 = 2A + 2B; and since we already know that B = 1, this gives A = 1.

At this point there are no more roots of p(x) to substitute (no more real roots, anyway!), but we can use the other linear equations (1),(2),(3),(4) now: substituting A=1 and B=1 into these equations, we get

(1() 1+C = 3

(2() –1+1–2C+D = –3

(3() 1+C–2D = 1

(4() –1+1+D = 1

so clearly C=2 and D=1.

Hence p(x)/q(x) = (3x3–3x2+x+1)/(x4–2x3+2x2–2x+1) equals

1/(x–1) + 1/(x–1)2 + (2x+1)/(x2+1), so the antiderivative of p(x)/q(x) is ( 1/(x–1) dx + ( 1/(x–1)2 dx + ( (2x+1)/(x2+1) dx.

Before we compute these three antiderivatives, let’s check to make sure that we’ve done the partial fractions decomposition correctly!

Check: 1/(x–1) + 1/(x–1)2 + (2x+1)/(x2+1)

= [(x–1)(x2+1) + (x2+1) + (2x+1)(x–1)2]/(x–1)2(x2+1)

= [(x3–x2+x–1) + (x2+1) + (2x3–3x2+1)]/(x–1)2(x2+1)

= (3x3–3x2+x+1)/ (x4–2x3+2x2–2x+1). (

Now at last we integrate:

First integral:

( 1/(x–1) dx =

..?..

( 1/u du (with u = x–1) =

..?..

ln |u| =

..?..

ln |x–1|.

Second integral:

( 1/(x–1)2 dx =

..?..

( 1/u2 du (with u = x–1) =

..?..

( u–2 du =

..?..

–u–1 =

..?..

–1/u = –1/(x–1)

Third integral:

( (2x+1)/(x2+1) dx splits into two pieces:

First piece: ( (2x)/(x2+1) dx =

..?..

( dv/v (with v = x2+1) =

..?..

ln |v| = ln |x2+1| = ln (x2+1)

Second piece: ( (1)/(x2+1) dx =

..?..

arctan x.

(Note that substituting v = x2+1, although good to try, doesn’t help us in this case; one needs to either recognize 1/(x2+1) as the derivative a function we’ve met before, or use a table of integrals, or use a computer algebra system.)

So we find that

( (3x3–3x2+x+1)/(x4–2x3+2x2–2x+1) dx equals

ln |x–1| – 1/(x–1) + (ln (x2+1) + arctan x).

And you can check this by …

..?..

..?..

differentiating (though I won’t give the details).

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