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COMEDK Exam 2012 Solved Paper
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© examsiri.com
Question :
147 of 180
Marks:
+1
,
-0
∫
x
x
(
1
+
log
‌
x
)
‌
dx
=
x
x
+
C
x
−
x
+
x
x
‌
log
‌
x
+
x
log
‌
x
+
x
Validate
Solution:
We have,
I
=
∫
x
x
(
1
+
l
o
g
x
)
d
x
Let
x
x
=
t
⇒
d
t
d
x
=
x
x
l
o
g
x
+
x
x
⇒
d
t
d
x
=
x
x
(
1
+
l
o
g
x
)
⇒
d
t
=
x
x
(
1
+
l
o
g
x
)
d
x
Now,
I
=
∫
d
t
=
t
+
C
=
x
x
+
C
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