partial fraction of 2x/(x-1) '2(x'2+1)
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Step-by-step explanation:
answer is
=
−
1
(
x
−
1
)
2
+
1
x
−
1
−
(
x
−
1
)
x
2
+
1
Explanation:
Perform the decomposition into partial fractions
x
2
−
2
x
−
1
(
x
−
1
)
2
(
x
2
+
1
)
=
A
(
x
−
1
)
2
+
B
x
−
1
+
C
x
+
D
x
2
+
1
=
A
(
x
2
+
1
)
+
B
(
x
−
1
)
(
x
2
+
1
)
+
(
C
x
+
D
)
(
x
−
1
)
2
(
x
−
1
)
2
(
x
2
+
1
)
The denominators are the same, compare the numerators
(
x
2
−
2
x
−
1
)
=
A
(
x
2
+
1
)
+
B
(
x
−
1
)
(
x
2
+
1
)
+
(
C
x
+
D
)
(
x
−
1
)
2
Let
x
=
1
−
2
=
2
A
,
A
=
−
1
Coefficients of
x
2
1
=
A
−
B
−
2
C
+
D
Coefficients of
x
−
2
=
B
+
C
−
2
D
and
−
1
=
A
−
B
+
D
,
⇒
,
−
B
+
D
=
0
,
B
=
D
1
=
−
1
−
B
−
2
C
+
B
,
⇒
,
2
C
=
−
2
,
⇒
,
C
=
−
1
−
2
=
B
−
1
−
2
B
,
⇒
,
B
=
1
⇒
,
D
=
1
Finally,
x
2
−
2
x
−
1
(
x
−
1
)
2
(
x
2
+
1
)
=
−
1
(
x
−
1
)
2
+
1
x
−
1
−
(
x
−
1
)
x
2
+
1
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