Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

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Partial Fractions Day 2 Chapter 7.4 April 3, 2007

Transcript of Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

Page 1: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

Partial Fractions Day 2

Chapter 7.4

April 3, 2007

Page 2: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

Integrate:x3 + x2 + 6x2 +1∫  dx

Chapter 7.4

April 3, 2007

Page 3: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

The Arctangent formula (also see day 9 notes)

The “new” formula:

When to use? If the polynomial in the denominator does not have real roots (b2-4ac < 0) then the integral is an arctangent, we complete the square and integrate….

For example:

1

u2 + a2 du=1atan−1 u

a⎛⎝⎜

⎞⎠⎟∫

1

2x2 +10x+13dx∫ =

11

22x + 5( )

2+ 1( )

dx∫ =21

2x + 5( )2

+ 1dx∫

u =2x+ 5du=2dx

=tan−1 2x + 5( )

Page 4: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

What if the polynomial has real roots?

That means we can factor the polynomial and “undo” the addition! To add fractions we find a common denominator and add: we’ll work the other way….

The denominator of our rational function factors into (t - 4)(t +1) So in our original “addition,” the fractions were of the form:

1

t 2 + 3t−4dt∫

a

b+cd

=ad+ cbbd

ad + cbbd

=ab

+cd

1

t 2 + 3t−4=

At−4

+Bt+1

Page 5: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

1

t 2 + 3t−4dt∫

1

t 2 + 3t−4=

At−4

+Bt+1

Page 6: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

Examples:

x−9x2 + 3x−10

dx∫ −2x − 6

x2 − 2x − 3∫  dx

Page 7: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

Each of these integrals involved linear factorsWhat if a factor is repeated?

For example:

The “x” factor is repeated, so in our original addition, we could have had each of the “reduced” fractions:

Clearing our denominators, we get:

x−1x3 + x2 dx∫ =

x −1

x2 x +1( )dx∫

x−1x2 x+1( )

=Ax

+Bx2 +

Cx+1

x−1=Ax(x+1) + B(x+1) +Cx2

Page 8: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

To Solve for A, B, and C, again we choose x carefully:x−1=Ax(x+1) + B(x+1) +Cx2

Page 9: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

Using this information, our original integral becomes:

x−1x3 + x2 dx∫ =

2x

+−1x2 +

−2x+1

⎛⎝⎜

⎞⎠⎟∫ dx

Page 10: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

Example:2x + 3x x−1( )2

dx∫

Page 11: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

We may also have expressions with factors of higher powers:

We apply the same concept as when there are linear factors, we undo the addition using REDUCED fractions.

2x2 + x−8x3 + 4x

dx∫

Page 12: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

We haveTo solve for B and C, we will match coefficients

From

2x2 + x−8 =A x2 + 4( ) + Bx+C( )x, A =−2

2x2 + x−8 =A x2 + 4( ) + Bx+C( )x

=Ax2 + 4A + Bx2 +Cx

=Ax2 + Bx2 +Cx + 4A

2x2 +1x−8 = A+ B( )x2 +Cx+ 4A

Page 13: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

The Integration:2x2 + x−8x x2 + 4( )

⎝⎜

⎠⎟∫ dx=

−2x

+4x+1x2 + 4

⎛⎝⎜

⎞⎠⎟∫ dx

Page 14: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.

Example:

x4

x4 −1∫  dx