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Standard XII
Mathematics
Question
If
∫
(
x
+
1
)
x
(
1
+
x
e
x
)
2
d
x
=
log
|
1
−
f
(
x
)
|
+
f
(
x
)
+
C
, then
f
(
x
)
=
1
x
+
e
x
1
1
+
x
e
x
1
(
1
+
x
e
x
)
2
1
(
x
+
e
x
)
2
A
1
(
x
+
e
x
)
2
B
1
x
+
e
x
C
1
(
1
+
x
e
x
)
2
D
1
1
+
x
e
x
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Solution
Verified by Toppr
Put
t
=
x
e
x
⇒
d
t
=
e
x
(
1
+
x
)
d
x
Thus,
∫
(
x
+
1
)
x
(
1
+
x
e
x
)
2
d
x
=
∫
e
x
(
1
+
x
)
d
x
x
e
x
(
1
+
x
e
x
)
2
=
∫
d
t
t
(
1
+
t
)
2
=
∫
(
1
t
−
1
1
+
t
−
1
(
1
+
t
)
2
)
d
t
=
log
|
t
|
−
log
|
1
+
t
|
+
1
1
+
t
+
c
=
log
∣
∣
∣
t
1
+
t
∣
∣
∣
+
1
1
+
t
+
c
=
log
∣
∣
∣
(
1
−
1
1
+
t
)
∣
∣
∣
+
1
1
+
t
+
c
∴
f
(
x
)
=
1
1
+
x
e
x
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