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Byju's Answer
Standard XII
Mathematics
First Fundamental Theorem of Calculus
∫ e x +1/x1+ ...
Question
∫
e
(
x
+
1
x
)
(
1
+
x
−
1
x
)
d
x
=
e
(
x
+
1
x
)
f
(
x
)
+
c
t
h
e
n
d
f
(
x
)
d
x
A
x
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B
0
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C
1
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D
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Solution
The correct option is
C
1
∫
e
x
+
1
x
+
x
e
x
+
1
x
(
1
−
1
x
2
)
d
x
=
x
e
x
+
1
x
+
c
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0
Similar questions
Q.
Obtain:
∫
(
e
x
−
1
)
(
e
x
−
1
)
√
e
x
+
1
+
e
−
x
d
x
Q.
Let
f
(
x
)
=
e
x
+
1
e
x
−
1
and
∫
1
0
e
x
+
1
e
x
−
1
.
x
d
x
=
λ
.
Then
∫
1
−
1
t
f
(
t
)
is equal to
Q.
Let
f
(
x
)
=
∫
e
x
x
d
x
and
∫
(
e
x
−
1
)
(
2
x
)
x
2
−
5
x
+
4
d
x
=
α
f
(
x
−
4
)
+
β
f
(
x
−
1
)
+
γ
, then
Q.
∫
x
e
x
(
1
+
x
)
2
d
x
=
Q.
If
l
n
=
∫
(
1
+
x
+
x
−
1
)
e
x
+
x
−
1
d
x
=
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First Fundamental Theorem of Calculus
Standard XII Mathematics
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