partial fractions day 2 chapter 7.4 april 3, 2006
TRANSCRIPT
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
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( )
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
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)
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
Examples:
x−9x2 + 3x−10
dx∫−2x − 6
x2 − 2x − 3∫ dx
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
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
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
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
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
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?
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:
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⎛⎝⎜
⎞⎠⎟
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
In groups, try:
x−9x2 + 3x−10
dx∫ 3x2 −56x+1∫ dx 2x−7
x3 + 2x∫ dx