2x - 1/ (x-1) (2x+3) in partial fractions
Answers
Answered by
5
Step-by-step explanation:
The answer is
=
−
1
(
x
−
1
)
3
−
3
(
x
−
1
)
2
−
3
x
−
1
+
3
x
−
2
Explanation:
Let's perform the decomposition into partial fractions
2
x
−
1
(
x
−
1
)
3
(
x
−
2
)
=
A
(
x
−
1
)
3
+
B
(
x
−
1
)
2
+
C
x
−
1
+
D
x
−
2
=
A
(
x
−
2
)
+
B
(
x
−
1
)
(
x
−
2
)
+
C
(
x
−
1
)
2
(
x
−
2
)
+
D
(
x
−
1
)
3
(
x
−
1
)
3
(
x
−
2
)
The denominators are the same, we compare the numerators
2
x
−
1
=
A
(
x
−
2
)
+
B
(
x
−
1
)
(
x
−
2
)
+
C
(
x
−
1
)
2
(
x
−
2
)
)
+
D
(
x
−
1
)
3
Let
x
=
1
,
⇒
,
1
=
−
A
,
⇒
,
A
=
−
1
Let
x
=
2
,
⇒
,
3
=
D
Coefficients of
x
3
0
=
C
+
D
,
⇒
,
C
=
−
D
=
−
3
Let
x
=
0
,
−
1
=
−
2
A
+
2
B
−
2
C
−
D
2
B
=
2
A
+
2
C
+
D
−
1
=
−
2
−
6
+
3
−
1
=
−
6
B
=
−
3
Therefore,
2
x
−
1
(
x
−
1
)
3
(
x
−
2
)
=
−
1
(
x
−
1
)
3
−
3
(
x
−
1
)
2
−
3
x
−
1
+
3
x
−
2
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