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KEAM 2015 Math Paper
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© examsnet.com
Question : 8
Total: 120
∫
4
e
x
2
e
x
−
5
e
−
x
d
x
=
4
log
|
e
x
−
5
|
+
C
1
4
log
|
e
2
x
−
5
|
+
C
log
|
e
x
−
5
e
−
x
|
+
C
4
log
|
e
2
x
−
5
|
+
C
log
|
2
e
2
x
−
5
|
+
C
Validate
Solution:
Let
I
=
∫
4
e
x
2
e
x
−
5
e
−
x
d
x
=
∫
4
e
x
2
e
x
−
5
e
x
d
x
=
∫
4
e
2
x
2
e
2
x
−
5
d
x
Put
2
e
2
x
−
5
=
t
⇒
4
e
2
x
d
x
=
d
t
∴
I
=
∫
d
t
t
=
log
|
t
|
+
C
⇒
I
=
log
|
2
e
2
x
−
5
|
+
C
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