partial fractions day 2 chapter 7.4 april 3, 2006

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Partial Fractions Day 2 Chapter 7.4 April 3, 2006

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Page 1: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

Partial Fractions Day 2

Chapter 7.4

April 3, 2006

Page 2: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

Integrate:x3 + x2 + 6x2 +1∫  dx

= x +1+−x + 5

x2 +1⎛⎝⎜

⎞⎠⎟∫  dx

= x + 1 +−x

x2 + 1+

5

x2 + 1⎛⎝⎜

⎞⎠⎟∫  dx

=x2

2+ x −

1

2ln x2 +1 + 5 tan−1 x

Page 3: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

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, 2006

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, 2006

What if the polynomial has real roots?

We now need to solve for A and B. Multiplying both sides by the denominator:

Choose t to be 1 and substituting into the equation above, we get 1 = 5B or B = 1/5

And with t = -4: 1 = -5A, or A = -1/5

1

t 2 + 3t−4dt∫

1

t 2 + 3t−4=

At+ 4

+Bt−1

1 =A(t−1)+ B(t+ 4)

Page 6: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

Integrating with the Partial Fraction Decomposition:

Our Rational function can be written as:

And the integral becomes:

Which integrates to:

1

t 2 + 3t−4dt∫

1

t 2 + 3t−4=

−15

t+ 4+

15t−1

1

t 2 + 3t−4dt∫ =−

15

1t+ 4

dt∫ +15

1t−1

dt∫

−1

5ln t + 4 +

1

5ln t −1 =

1

5lnt −1

t + 4

Page 7: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

Examples:

x−9x2 + 3x−10

dx∫−2x − 6

x2 − 2x − 3∫  dx

Page 8: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

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 9: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

To Solve for A, B, and C, again we choose x carefully:

If we let x = 0, we can solve for B,

Next, if x=-1, we can solve for C,

We still need “A.” But because we know B and C, we can choose any x and use the known values to solve for A. Let x = 1

x−1=Ax(x+1) + B(x+1) +Cx2

0 −1=A0(0 +1) + B(0 +1) +C02 ⇒ B = −1

−1−1 = A −1( )(−1+1) + B(−1+1) +C −1( )2 ⇒ C = −2

1−1=A 1( )(1+1) + B(1+1) +C 1( )2

0 =2A+ 2B+C

0 =2A+ 2 −1( ) + −2( ) ⇒ A = 2

Page 10: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

Using this information, our original integral becomes:

We can integrate each term resulting in:

Another form of the answer might be:

x−1x3 + x2 dx∫ =

2x

+−1x2 +

−2x+1

⎛⎝⎜

⎞⎠⎟∫ dx

x−1x3 + x2 dx∫ =2 ln x +

1x

−2 ln x+1

=2 lnx

x +1+

1

x

=lnx2

x +1( )2 +

1

x

Page 11: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

Example:

Undoing the addition, we get:

Solving for A, B and C, we arrive at:

Integrating, we get:

2x + 3x x−1( )2

dx∫

2x + 3x x−1( )2

=Ax

+Bx−1

+C

x−1( )2

2x + 3x x−1( )2

=3x

+−3x−1

+5

x−1( )2

2x + 3x x−1( )2∫ dx=

3x

+−3x−1

+5

x−1( )2⎛

⎝⎜

⎠⎟∫ dx

=3ln x − 3ln x −1 −5

x −1=3ln

x

x −1−

5

x −1=ln

x3

x −1( )3 −

5

x −1

Page 12: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

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.

Clearing the denominator, we get:

Letting x = 0, we can solve for A:

Solving for B and C, we need a new strategy………

2x2 + x−8x3 + 4x

dx∫

2x2 + x−8x x2 + 4( )

=Ax

+?

x2 + 4

The highest power on top must be less than the power on the bottom.

2x2 + x−8x x2 + 4( )

=Ax

+Bx+Cx2 + 4

2x2 + x−8 =A x2 + 4( ) + Bx+C( )x

2 0( )2 + 0( )−8 =A 0( )2 + 4( ) + B 0( ) +C( ) 0( ) ⇒ A = −2

Page 13: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

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

From

Matching the coefficients of both sides of the equation, term by term, we get:

Returning to our original integral, we get that:

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 + x−8 = A+ B( )x2 +Cx+ 4A

2 = A+ B( )1 =C

−8 = 4A

⇒ 2 = −2 + B ⇒ 4 = B

2x2 + x−8x x2 + 4( )

⎝⎜

⎠⎟∫ dx=

−2x

+4x+1x2 + 4

⎛⎝⎜

⎞⎠⎟∫ dx

Integrating the -2/x is clear, but what about the second term?

Page 14: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

The Integration:2x2 + x−8x x2 + 4( )

⎝⎜

⎠⎟∫ dx=

−2x

+4x+1x2 + 4

⎛⎝⎜

⎞⎠⎟∫ dx

−2

xdx +∫

4x +1

x2 + 4⎛⎝⎜

⎞⎠⎟∫ dx =−2 ln x +

4x +1

x2 + 4⎛⎝⎜

⎞⎠⎟∫ dx

We also will need to split the remaining integral.

=−2 ln x +4x

x2 + 4⎛⎝⎜

⎞⎠⎟∫ dx +

1

x2 + 4⎛⎝⎜

⎞⎠⎟∫ dxThe first integral can be

solved using u-substitution

Splitting the integral, we get:

Page 15: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

The Integration:2x2 + x−8x x2 + 4( )

⎝⎜

⎠⎟∫ dx=

−2x

+4x+1x2 + 4

⎛⎝⎜

⎞⎠⎟∫ dx

=−2 ln x +4x

x2 + 4⎛⎝⎜

⎞⎠⎟∫ dx +

1

x2 + 4⎛⎝⎜

⎞⎠⎟∫ dx

The first integral can be solved using u-substitution

u =x2 + 4du=2xdx 2

2x

x2 + 4⎛⎝⎜

⎞⎠⎟∫ dx

2du

u⎛⎝⎜

⎞⎠⎟∫

2 ln x2 + 4

The second integral is an arctangent

+1

2tan−1 x

2⎛⎝⎜

⎞⎠⎟

Final Solution: =−2 ln x +

=2 lnx2 + 4

x+

1

2tan−1 x

2⎛⎝⎜

⎞⎠⎟=ln

x2 + 4( )2

x2+

1

2tan−1 x

2⎛⎝⎜

⎞⎠⎟

Page 16: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

More Examples:

x4

x4 −1∫  dx = 1+1

x4 −1⎛⎝⎜

⎞⎠⎟∫  dx = 1+

1

4x −1

+−

1

4x +1

+−

1

2x2 +1

⎜⎜⎜

⎟⎟⎟

∫  dx

1 =A x+1( ) x2 +1( ) + B x−1( ) x2 +1( ) + Cx+D( ) x−1( ) x+1( )

1

x4 −1=

Ax−1

+Bx+1

+Cx+Dx2 +1

Page 17: Partial Fractions Day 2 Chapter 7.4 April 3, 2006

In groups, try:

x−9x2 + 3x−10

dx∫ 3x2 −56x+1∫  dx 2x−7

x3 + 2x∫  dx