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IIT JEE Advanced 2008 Paper 2
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Section:
Mathematics
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© examsnet.com
Question : 48
Total: 66
Let I =
∫
e
x
e
4
x
+
e
2
x
+
1
dx, J =
∫
e
−
x
e
−
4
x
+
e
−
2
x
+
1
dx. Then, for an arbitrary constant C, the value of J - 1 equals
[JEE Adv 2008 P2]
1
2
‌
log
(
e
4
x
−
e
2
x
+
1
e
4
x
+
e
2
x
+
1
)
+ C
1
2
‌
log
(
e
2
x
+
e
x
+
1
e
2
x
−
e
x
+
1
)
+ C
1
2
‌
log
(
e
2
x
−
e
x
+
1
e
2
x
+
e
x
+
1
)
+ C
1
2
‌
log
(
e
4
x
+
e
2
x
+
1
e
4
x
−
e
2
x
+
1
)
+ C
Validate
Solution:
We have
I =
∫
e
x
e
4
x
+
e
2
x
+
1
dx
J =
∫
e
−
x
e
−
4
x
+
e
−
2
x
+
1
dx
=
∫
e
3
x
d
x
e
4
x
+
e
2
x
+
1
Now , J - I =
∫
(
e
2
x
−
1
e
x
d
x
e
4
x
+
e
2
x
+
1
Substituting
e
x
= t, we get
J - I =
∫
(
t
2
−
1
)
d
t
t
4
+
t
2
+
1
=
∫
(
1
−
1
t
2
)
d
t
(
t
+
1
t
)
2
+
1
=
1
2
‌
log
(
e
x
+
e
−
x
−
1
e
x
+
e
−
x
+
1
)
+
c
=
1
2
‌
log
(
e
2
x
+
e
x
+
1
e
2
x
+
e
x
+
1
)
+
c
© examsnet.com
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