evaluate the integral ∫ (x^2-2x-1/(x+1)^2(x^2+1)) dx
Answers
Answer:
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Step-by-step explanation:
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Step-by-step explanation:
Let us first convert
x
2
−
2
x
−
1
(
x
−
1
)
2
(
x
2
+
1
)
to partial fractions. For this let
x
2
−
2
x
−
1
(
x
−
1
)
2
(
x
2
+
1
)
⇔
A
x
−
1
+
B
(
x
−
1
)
2
+
C
x
+
D
x
2
+
1
or
x
2
−
2
x
−
1
(
x
−
1
)
2
(
x
2
+
1
)
=
A
(
x
−
1
)
(
x
2
+
1
)
+
B
(
x
2
+
1
)
+
(
C
x
+
D
)
(
x
−
1
)
2
(
x
−
1
)
2
(
x
2
+
1
)
or
(
x
2
−
2
x
−
1
)
=
A
(
x
−
1
)
(
x
2
+
1
)
+
B
(
x
2
+
1
)
+
(
C
x
+
D
)
(
x
−
1
)
2
Putting
x
=
1
in this we get
2
B
=
−
2
or
B
=
−
1
Comparing coefficient of highest degree i.e.
x
3
, we get
A
+
C
=
0
,
comparing constant term we get
−
A
+
B
+
D
=
−
1
or
−
A
+
D
=
0
i.e.
A
=
D
and comparing coefficient of
x
,
−
2
=
A
−
2
D
+
C
i.e.
C
−
A
=
−
2
as
A
+
C
=
0
, we get
C
=
−
1
and
A
=
1
and hence
(x^2-2x-1)/((x-1)^2(x^2+1))=1/(x-1)-1/(x-1)^2-(x-1)/(x^2+1)
x
2
−
2
x
−
1
(
x
−
1
)
2
(
x
2
+
1
)
=
1
x
−
1
−
1
(
x
−
1
)
2
−
x
−
1
x
2
+
1
Hence int(x^2-2x-1)/((x-1)^2(x^2+1))dx
∫
x
2
−
2
x
−
1
(
x
−
1
)
2
(
x
2
+
1
)
d
x
= intdx/(x-1)-intdx/(x-1)^2-int(x-1)/(x^2+1)dx
∫
d
x
x
−
1
−
∫
d
x
(
x
−
1
)
2
−
∫
x
−
1
x
2
+
1
d
x
= ln|x-1|+1/(x-1)-1/2int(2x)/(x^2+1)dx+intdx/(x^2+1)
ln
|
x
−
1
|
+
1
x
−
1
−
1
2
∫
2
x
x
2
+
1
d
x
+
∫
d
x
x
2
+
1
= ln|x-1|+1/(x-1)-1/2ln(x^2+1)+tan^(-1)x+c
ln
|
x
−
1
|
+
1
x
−
1
−
1
2
ln
(
x
2
+
1
)
+
tan
−
1
x
+
c