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UPSC NDA Math Model Paper 2
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© examsnet.com
Question : 6
Total: 120
The value of
∫
e
x
e
2
x
−
4
d
x
will be ______, where
C
is an arbitrary constant.
1
2
log
|
e
x
+
1
e
x
−
1
|
+
C
1
3
log
|
2
e
x
−
1
2
e
x
+
1
|
+
C
1
4
log
|
e
x
−
2
e
x
+
2
|
+
C
1
2
log
|
e
2
x
+
2
e
2
x
−
2
|
+
C
Validate
Solution:
∫
e
x
e
2
x
−
4
d
x
∫
e
x
(
e
x
)
2
−
(
2
)
2
d
x
let
t
=
e
x
d
t
=
e
x
d
x
∫
d
t
(
t
)
2
−
(
2
)
2
From the standard integral:
∫
d
t
(
t
)
2
−
(
2
)
2
=
1
4
log
|
t
−
a
t
+
a
|
+
C
Put
t
=
e
x
in the above equation, we get:
∫
e
x
e
2
x
−
4
d
x
=
1
4
log
|
e
x
−
2
e
x
+
2
|
+
C
Note:
Some important formulas of integration are:
∫
d
x
a
2
−
x
2
=
1
2
a
log
|
a
+
x
a
−
x
|
+
C
,
x
<
a
∫
d
x
√
a
2
−
x
2
=
sin
−
1
(
x
a
)
+
C
∫
d
x
√
x
2
+
a
2
=
log
(
x
+
√
a
2
+
x
2
)
+
C
© examsnet.com
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