Partial Fractions Day 2 Chapter 7.4 April 3, 2007.
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Transcript of Partial Fractions Day 2 Chapter 7.4 April 3, 2007.
![Page 1: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/1.jpg)
Partial Fractions Day 2
Chapter 7.4
April 3, 2007
![Page 2: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/2.jpg)
Integrate:x3 + x2 + 6x2 +1∫ dx
Chapter 7.4
April 3, 2007
![Page 3: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/3.jpg)
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.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/4.jpg)
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.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/5.jpg)
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.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/6.jpg)
Examples:
x−9x2 + 3x−10
dx∫ −2x − 6
x2 − 2x − 3∫ dx
![Page 7: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/7.jpg)
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.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/8.jpg)
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.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/9.jpg)
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.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/10.jpg)
Example:2x + 3x x−1( )2
dx∫
![Page 11: Partial Fractions Day 2 Chapter 7.4 April 3, 2007.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/11.jpg)
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.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/12.jpg)
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.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/13.jpg)
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.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0141a28abf838ccd26e/html5/thumbnails/14.jpg)
Example:
x4
x4 −1∫ dx