If the derivative of any function ln (kx) for positive constant k is always 1/x . . .?

Question

When integrating 1/x couldn't you make the antiderivative anything of the form ln (kx)? For example, you could say the antiderivative of 1/x is ln x , ln 5x, ln 12x, etc. But these would give you different results when integrating . . . ? ? ? ? ? ?

Answer 1

The anit-derivative of 1/x is not ln x, it is:
ln x + C
Imagine you pick that C in a special way so that C is ln K, where K is some other constant. now:
ln x + ln K = ln (K)(x) by the properties of logs.

That darn constant!

I once had a long argument with my math teacher in high school about an antiderivative where we got slightly different answers. As I was falling asleep that night I shot up in my bed...Our answers were off by only a constant! That's the only time I've ever solved a math problem subconciously. It was surreal.

Answer 2

You are forgeting your + C at the end of the integral. Integrating (1/x) gives you ln x + C.
Perhaps the C is ln of 5
which would give you ln x + ln 5 or ln 5x

Answer 3

Absolutely.

Integral (1/x dx) =

ln|x| + C

But C is the same as ln(e^c), as ln and e are inverses of each other, which means we get

ln|x| + ln(e^c)

Combining these two using the log property that
log[base b](a) + log[base b](c) = log[base b](ac), we get

ln| e^c (x) |

But e^c is just a constant obtained from a constant, so why not just call it a constant, k?

ln|kx| is our answer.

Answer 4

y = c dy/dx = 0 dy/dx is slightly identically 0, on a similar time as c = a million For the 2nd. f(x) = kx^a million/2 -lnx f' = a million/2 * kx^-a million/2 - a million/x f''(x) = -a million/4 * ok * x-(3/2) + a million/x^2 on the factor of inflexion, f'' = 0 x^a million/2 = 4/ok So the y cordinate ought to be 0 for this to be on the x axis, so f(4/ok) = 4 - ln(4/ok) = 0 ok = 4 / e^4

Answer 5

Sure..
the antiderivative of 1/x is ln(x) +c
if you let it be ln(kx) then.
ln(kx) = ln(k) + ln(x)
ln(k) is just a constant, or part of c