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JEE Main Math Class 12 Integrals Part 3 Questions
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© examsnet.com
Question : 8
Total: 100
For
α
,
β
,
γ
,
δ
∈
N
, if
∫
(
(
x
e
)
2
x
+
(
e
x
)
2
x
)
log
e
xdx
=
1
α
(
x
e
)
β
x
−
1
γ
(
e
x
)
δ
x
+
C
, where
e
=
∞
∑
n
=
0
1
n
!
and
C
is constant of integration, then
α
+
2
β
+
3
γ
−
4
δ
is equal to
[10-Apr-2023 shift 2]
4
-4
-8
1
Validate
Solution:
x
=
e
ln
x
∫
(
(
x
e
)
2
x
+
(
e
x
)
2
x
)
log
e
x
dx
=
∫
[
e
2
(
x
ln
x
−
x
)
+
e
−
2
(
x
ln
x
−
x
)
]
ln
x
dx
x
ln
x
−
x
=
t
ln
x
⋅
dx
=
d
t
∫
(
e
2
t
+
e
−
2
t
)
dt
e
2
t
2
−
e
−
2
t
2
+
C
=
1
2
(
x
e
)
2
x
−
1
2
(
e
x
)
2
x
+
C
α
=
B
=
γ
=
δ
=
2
α
+
2
B
+
3
γ
−
4
δ
=
4
© examsnet.com
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