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KEAM 2015 Math Paper
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© examsnet.com
Question : 12
Total: 120
∫
1
x
log
x
2
d
x
=
1
2
log
|
log
x
2
|
+
C
log
|
log
x
2
|
+
C
2
log
|
log
x
2
|
+
C
4
log
|
log
x
2
|
+
C
1
4
log
|
log
x
2
|
+
C
Validate
Solution:
Let
I
=
∫
1
x
log
x
2
d
x
Put
log
x
2
=
t
⇒
1
x
2
⋅
2
x
d
x
=
d
t
⇒
2
x
d
x
=
d
t
∴
I
=
1
2
∫
1
t
d
t
=
1
2
log
|
t
|
+
C
=
1
2
log
|
log
x
2
|
+
C
© examsnet.com
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