Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

University of Toronto Scarborough
MATA36
Calculus II for Physical Sciences
Text Book: Calculus Early Transcendentals 9th Edition by James
Stewart, Daniel Clegg, Saleem Watson
Partial fraction decomposition is the breaking down of a fraction into multiple fractions where
the numerator is a constant
...
This helps in integrating the original fraction
...

The first step is to factorize the denominator, here, the denominator is already factorized
...
To do that, multiply both
sides of the equation (decomposed and undecomposed) by the factorized denominator
...
For
example, if we input, x=3, then B (3-3) = 0, leaving only the constant A
...

Input x = 3,
1 = 𝐴(3 + 5) + 𝐡(3 βˆ’ 3)
1 = 8𝐴
𝐴=

1
8

Input x = -5
1 = 𝐴(βˆ’5 + 5) + 𝐡(βˆ’5 βˆ’ 3)
1 = βˆ’8𝐡
𝐡= βˆ’

1
8
1

Remembering ∫ π‘₯ 𝑑π‘₯ = 𝑙𝑛|π‘₯| + 𝐢, and using the substitution method,
The decomposed equation we get and its integration is as follows,
∫

1
1
1
1
1
= ∫
βˆ’
=∫
βˆ’βˆ«
(π‘₯ βˆ’ 3)(π‘₯ + 5)
8(π‘₯ βˆ’ 3) 8(π‘₯ + 5)
8(π‘₯ βˆ’ 3)
8(π‘₯ + 5)
1
1
= 𝑙𝑛|π‘₯ βˆ’ 3| βˆ’ 𝑙𝑛|π‘₯ + 5| + 𝐢
8
8

______________________________________________________________________________
∫

π‘₯2 + 4
𝑑π‘₯
π‘₯ 3 βˆ’ 3π‘₯ 2 + 2π‘₯

Here, the denominator is not factorized, so we have to do that
...


π‘₯2 + 4
𝐴
𝐡
𝐢
π‘₯(π‘₯ βˆ’ 1)(π‘₯ βˆ’ 2) = +
+
(π‘₯(π‘₯ βˆ’ 1)(π‘₯ βˆ’ 2))
π‘₯(π‘₯ βˆ’ 1)(π‘₯ βˆ’ 2)
π‘₯ π‘₯βˆ’1 π‘₯βˆ’2
π‘₯ 2 + 4 = 𝐴(π‘₯ βˆ’ 1)(π‘₯ βˆ’ 2) + 𝐡(π‘₯)(π‘₯ βˆ’ 2) + 𝐢(π‘₯)(π‘₯ βˆ’ 1)
π‘₯ 2 + 4 = 𝐴(π‘₯ 2 βˆ’ 2π‘₯ βˆ’ π‘₯ + 2) + 𝐡(π‘₯ 2 βˆ’ 2π‘₯) + 𝐢 (π‘₯ 2 βˆ’ π‘₯)

Notice that here, the strategy of inputting certain x-values won’t work
...

𝐴(π‘₯ 2 βˆ’ 2π‘₯ βˆ’ π‘₯ + 2) + 𝐡(π‘₯ 2 βˆ’ 2π‘₯) + 𝐢(π‘₯ 2 βˆ’ π‘₯)
= 𝐴π‘₯ 2 βˆ’ 2𝐴π‘₯ βˆ’ 𝐴π‘₯ + 2𝐴 + 𝐡π‘₯ 2 βˆ’ 2𝐡π‘₯ + 𝐢π‘₯ 2 βˆ’ 𝐢π‘₯

Now, arrange everything in terms of the power of x
...

2A=4
A=2
Adding (1) and (2) we get,
-2A-B=1
Input A=2
-2(2)-B=1
-B=5
B=-5
From (1),
A+B+C=1
2 + (-5) + C=1
C=4
2
5
4
βˆ’
+
π‘₯ π‘₯βˆ’1 π‘₯βˆ’2
2
5
4
2
5
4
∫ βˆ’
+
𝑑π‘₯ = ∫ 𝑑π‘₯ βˆ’ ∫
𝑑π‘₯ + ∫
𝑑π‘₯
π‘₯ π‘₯βˆ’1 π‘₯βˆ’2
π‘₯
π‘₯βˆ’1
π‘₯βˆ’2
= 2𝑙𝑛|π‘₯| βˆ’ 5𝑙𝑛|π‘₯ βˆ’ 1| + 4𝑙𝑛|π‘₯ βˆ’ 2| + 𝐢
______________________________________________________________________________

∫

π‘₯ βˆ’ 12
𝑑π‘₯
π‘₯ 2 βˆ’ 4π‘₯

Let’s factorize the denominator
...

π‘₯ βˆ’ 12
𝐴
𝐡
π‘₯(π‘₯ βˆ’ 4) = ( +
)π‘₯(π‘₯ βˆ’ 4)
π‘₯(π‘₯ βˆ’ 4)
π‘₯ π‘

---