# Integration of (1/kx)

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#1
If you want to do this:
where k is a constant, you can either take that as: or , which respectively integrate to and . But obviously they're different so what is the explanation???????
1
13 years ago
#2
You are forgetting the constant that comes out as well (the + C) which can combine with whatever is inside the ln
1
13 years ago
#3
(Original post by mrsmann)
If you want to do this:
where k is a constant, you can either take that as: or , which respectively integrate to and . But obviously they're different so what is the explanation???????
1
#4
Ok thanks for the replies.

+c sucks
0
13 years ago
#5
(Original post by mrsmann)
Ok thanks for the replies.

+c sucks
+c > you.
0
#6
(Original post by generalebriety)
mrsmann > +c > me.
Fixed!
0
13 years ago
#7
(Original post by mrsmann)
Fixed!
Ah yes. I forgot to carry the 1.
0
#8
I don't get it
0
13 years ago
#9
1/k lnk is just a constant....
0
1 year ago
#10
(Original post by Sudarshan stud)
U can also do like this right take (kx) to numerator it will become (kx)pow -1, then integrate, it becomes k×((x)pow-2/-2)..... Then which is right.........????????
This thread is 12 years old. When you integrate you add 1 to the power. You have subtracted 1, but it doesn't work when the power is - 1. I suggest you make a new thread instead.
0
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