## Thursday, May 3, 2012

### Calculus Revisited #14: Integration - Decomposition of Fractions

Welcome to Part 14 of our 21 part of our Calculus Revisited. Today's blog entry involves integration using decomposition of fractions.

Decomposition of Fractions

Here is the general strategy.

If the denominator has (x - c)^r then:

α / (x - c)^r = A1 / (x - c) + A2 / (x - c)^2 + ... + Ar / (x - c)^r

If the denominator has (x^2 + cx + d)^r then:

β / (x^2 + cx + d) = (A1x + B1) / (x^2 + cx + d) + (A2 x + B2) / (x^2 + cx + d)^2 + ... + (Ar x + Br) / (x^2 + cx + d)^r

Here are some examples.

Problems

Sometimes polynomial division is all you need.

1. ∫ (8x - 3) / (4x + 1) dx

First execute the division (8x - 3) / (4x + 1).

(8x - 3) / (4x + 1) = 2 - 5 / (4x + 1)

So

∫ (8x - 3) / (4x + 1) dx
= ∫ 2 - 5 / (4x + 1) dx
= 2x - 5/4 * ln(4x + 1)

Final: ∫ (8x - 3) / (4x + 1) dx = 2x - 5/4 * ln(4x+ 1) + C

Now here are two basic examples of decomposition of fractions.

2. ∫ dx / ((x + 1)(2x + 3))

Then:

A / (x + 1) + B / (2x + 3 ) = 1 / ((x + 1)(2x + 3))

A (2x + 3) + B (x + 1) = 1

2A(x) + 3A + B(x) + B = 1 + 0(x)

Look at the coefficients and match them.

Constant: 3A + B = 1
Coef. of x: 2A + B = 0

The next task is to solve for A and B.

3A + B = 1
2A + B = 0

Subtract the second equation from the first equation

A = 1

Find B

Using the first equation:
3(1) + B = 1
B = 1 -3
B = -2

Using the second equation:
2(1) + B = 0
2 + B = 0
B = -2

Conclude A = 1, B = -2

Back to our integral:

∫ dx / ((x + 1)(2x + 3))
= ∫ A / (x + 1) + B / (2x + 3) dx
= ∫ 1 / (x + 1) - 2 / (2x + 3) dx
= ln(x + 1) - ln(2x + 3)
= ln( (x + 1) / (2x + 3) )

Final: ∫ dx / ((x + 1)(2x + 3)) = = ln( (x + 1) / (2x + 3) ) + C

3. ∫ (2x + 1) / ( (x^2 + 1) (x - 1) ) dx

Decomposing:

(2x + 1) / ( (x^2 + 1) (x - 1) ) = (Ax + B) / (x^2 + 1) + C / (x - 1)

Then:

(Ax + B) (x - 1) + C (x^2 + 1) = 2x + 1
A(x^2) - A(x) + B(x) - B + C(x^2) + C = 2x + 1 + 0(x^2)
(A + C)x^2 + (-A + B)x + (-B + C) = (0)x^2 + (2)x + 1

Matching up terms of x^2, x, and the constant:
A + C = 0
-A + B = 2
-B + C = 1

From the first equation, C = -A (alternatively A = -C)

Using the second and third equations:
-A + B = 2
-B + C = 1

-A + B = 2
-B + (-A) = 1

-A + B = 2
-A - B = 1

Add the two equations together to get

-2A = 3
A = -3/2

Then: -A - B = 1
-(-3/2) - B = 1
- B = 1 - 3/2
B = 3/2 - 1
B = 1/2

So A = -3/2, B = 1/2, and C = 3/2

And:
∫ (2x + 1) / ( (x^2 + 1) (x - 1) ) dx
= ∫ (Ax + B) / (x^2 + 1) + C / (x - 1) dx
= ∫ ((-3/2)x + 1/2) / (x^2 + 1) + (3/2) / (x - 1) dx
= ∫ (-3/2)x / (x^2 + 1) + (1/2) / (x^2 + 1) + (3/2) / (x - 1) dx
= (-3/4) ln(x^2 + 1) + (1/2) atan x + (3/2) ln (x - 1)

Final:
∫ (2x + 1) / ( (x^2 + 1) (x - 1) ) dx = (-3/4) ln(x^2 + 1) + (1/2) atan x + (3/2) ln (x - 1) + C

That concludes our integration techniques section. Next time we move to infinite limits.

Until next time, take care,

Eddie

This blog is property of Edward Shore. © 2012