integration of 1/x(1-x)(1+x)
Answers
Answered by
1
Answer:
The answer is
=
−
ln
(
∣
x
∣
)
+
1
2
ln
(
∣
x
+
1
∣
)
+
ln
(
∣
x
−
1
∣
)
+
C
Explanation:
Let's do the `partial fraction decomposition
1
(
x
)
(
x
+
1
)
(
x
−
1
)
=
A
x
+
B
x
+
1
+
C
x
−
1
=
A
(
x
+
1
)
(
x
−
1
)
+
B
(
x
)
(
x
−
1
)
+
C
(
x
)
(
x
+
1
)
(
x
)
(
x
+
1
)
(
x
−
1
)
Therefore,
1
=
A
(
x
+
1
)
(
x
−
1
)
+
B
(
x
)
(
x
−
1
)
+
C
(
x
)
(
x
+
1
)
Let
x
=
0
,
⇒
,
1
=
−
A
,
⇒
,
A
=
−
1
Let
x
=
1
,
⇒
,
1
=
2
C
,
⇒
,
C
=
1
2
Let
x
=
−
1
,
⇒
,
1
=
2
B
,
⇒
,
B
=
1
2
So,
1
(
x
)
(
x
+
1
)
(
x
−
1
)
=
−
1
x
+
1
2
x
+
1
+
1
2
x
−
1
∫
d
x
(
x
)
(
x
+
1
)
(
x
−
1
)
=
∫
−
d
x
x
+
1
2
∫
d
x
x
+
1
+
1
2
∫
d
x
x
−
1
=
−
ln
(
∣
x
∣
)
+
1
2
ln
(
∣
x
+
1
∣
)
+
ln
(
∣
x
−
1
∣
)
+
C
Similar questions