mrsmann
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#1
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#1
If you want to do this:
\int{\frac{1}{kx}} where k is a constant, you can either take that as: \int{\frac{1}{k}*\frac{1}{x}} or \int{\frac{1}{kx}}, which respectively integrate to \frac{1}{k}lnx and \frac{1}{k}lnkx. But obviously they're different so what is the explanation???????
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Trangulor
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You are forgetting the constant that comes out as well (the + C) which can combine with whatever is inside the ln
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khaixiang
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#3
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(Original post by mrsmann)
If you want to do this:
\int{\frac{1}{kx}} where k is a constant, you can either take that as: \int{\frac{1}{k}*\frac{1}{x}} or \int{\frac{1}{kx}}, which respectively integrate to \frac{1}{k}lnx and \frac{1}{k}lnkx. But obviously they're different so what is the explanation???????
\frac{1}{k}lnkx=\frac{1}{k}\ln{k  }+\frac{\ln{x}}{k}=\frac{\ln{x}}  {k}+c
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mrsmann
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Ok thanks for the replies.

+c sucks
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generalebriety
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(Original post by mrsmann)
Ok thanks for the replies.

+c sucks
+c > you.
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mrsmann
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(Original post by generalebriety)
mrsmann > +c > me.
Fixed!
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generalebriety
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(Original post by mrsmann)
Fixed!
Ah yes. I forgot to carry the 1.
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mrsmann
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#8
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I don't get it
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insparato
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1/k lnk is just a constant....
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NotNotBatman
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(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.
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