Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
For 'c' is an arbitrary constant, ∫ elog x dx =
Q. For
′
c
′
is an arbitrary constant,
∫
e
l
o
g
x
d
x
=
1848
227
J & K CET
J & K CET 2019
Report Error
A
x
/2
+
c
B
e
l
o
gx
+
c
C
x
2
/2
+
c
D
x
+
c
Solution:
We have,
∫
e
l
o
g
x
d
x
=
∫
x
d
x
=
2
x
2
+
c