How to partial fractions!
(focus on the setups)
ideally, we like/want to integrate the
following rational functions…
1
1
1
1
−1 ⎛ x ⎞
⌠
⌠
dx = ln ax + b +C ⎮ 2 2 dx = tan ⎜ ⎟ +C
⎮
a
a
⎝a⎠
⌡ ax + b
⌡ x +a
completing the square might be
necessary in order to get here
do u sub
1
x
⌠
u = ax + b
do u sub
⌠
dx
dx
⎮ 2
⎮
2
n
then
reverse
u
=
ax
+b
⌡ ax + b
⌡ (ax + b )
n ≠ 1 power rule
1
1
2
=
(ax + b )1−n + C
=
ln
ax
+ b +C
a (1− n )
2a
unfortunately we don’t always
get what we like or what we
want…. : (
but don’t worry, partial
fraction decomposition (PFD)
comes to our rescue!
Case 0: Polynomial Long Division
Make sure the degree of the
numerator is less than the degree
of the denominator.
If not, then perform polynomial
long division.
Case 1: Distinct Linear Factors
1
1
A
B
[ex1.] 2
=
=
+
x −7x +10 (x −2)(x −5) x −2 x −5
A
B
2x +1
2x +1
=
=
+
[ex2.] 2
x −7x +10 (x −2)(x −5) x −2 x −5
A
B
C
3x −5
=
+
+
[ex3.]
(x −1)(x −3)(x −5) x −1 x −3 x −5
Irreducible Quadratic
Case 2: Factors
4x − 9x +2
A
Bx + C
[ex4.]
=
+ 2
2
(x +3)(x +4) x +3 x +4
2
x +5x +26
x +5x +26
=
[ex5.] 3
2
2
x (x +4x +13)
x +4x +13x
2
2
A
Bx + C
= + 2
x x +4x +13
Case 3: Repeating Factors
2x −5
2x −5 Ax + B C
Ax B
C
=
+
= 2 + 2+
[ex6.] 3 2 = 2
2
x +1 x
x x +1
x +x
x (x +1) x
C
A B
= 1 + 2+
x x x +1
A
B
C
D
4
x
−1
[ex7.]
= +
+
+
1
2
3
3
x (x +2) (x +2) (x +2)
x (x +2)
x +5
A Bx + C Dx + E
= + 2
+ 2
[ex8.]
2
2
1
2
x (x +4) x (x +4) (x +4)
**I will never ask this kind of
question on my calc exams**
You Try
2x −1
(Q1.)
(x −1)(x −2)(x −3)
2x 2 +8x +5
(Q 2.) 2
x +4x +13
3x 2 −3x +8
(Q 3.) 3
x −3x 2 +4x −12
x 2 +7x −3
(Q 4.)
(2x +1)(x −2)2
Thanks for watching!