resolve x square upon in (1 - X) into (1 + x square) square into partial fraction
Answers
Answered by
0
Answer:
The answer is
=
−
1
4
x
+
1
+
1
2
(
x
−
1
)
2
+
1
4
x
−
1
Explanation:
Let's factorise the denominator
x
2
−
1
=
(
x
+
1
)
(
x
−
1
)
Therefore,
x
(
x
2
−
1
)
(
x
−
1
)
=
x
(
x
+
1
)
(
x
−
1
)
(
x
−
1
)
=
x
(
x
+
1
)
(
x
−
1
)
2
Let's perform the decomposition into partial fractions
x
(
x
2
−
1
)
(
x
−
1
)
=
A
x
+
1
+
B
(
x
−
1
)
2
+
C
x
−
1
=
A
(
x
−
1
)
2
+
B
(
x
+
1
)
+
C
(
x
−
1
)
(
x
+
1
)
(
x
+
1
)
(
x
−
1
)
2
The denominators are the same, we compare the numerators
x
=
A
(
x
−
1
)
2
+
B
(
x
+
1
)
+
C
(
x
−
1
)
(
x
+
1
)
Let
x
=
−
1
,
⇒
,
−
1
=
4
A
,
⇒
,
A
=
−
1
4
Let
x
=
1
,
⇒
,
1
=
2
B
,
⇒
,
B
=
1
2
Coefficients of
x
2
0
=
A
+
C
,
⇒
,
C
=
−
A
=
1
4
Therefore,
x
(
x
2
−
1
)
(
x
−
1
)
=
−
1
4
x
+
1
+
1
2
(
x
−
1
)
2
+
1
4
x
−
1
Answered by
4
Answer:
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