math anomaly? tell me what's wrong!?

Question

integral (1/x dx) = ln x
but if I use integration by parts..
integral (1/x dx) where u= 1/x; dV=dx; dU =-(1/x^2)dx; V=x
it becomes:
integral (1/x dx)= (1/x)*x - int (x * (-1/x^2) dx
so
int (1/x dx)= 1 + int (1/x dx)
thus ln x = 1 + ln x....... !? if u substract both side by ln x it becomes 0=1 !!!!???
tell me what's wrong..
i don't think it's a matter of the constant...

Answer 1

well, the root cause of your trouble is your assumption that

integral (1/x dx) = ln x

It is, in fact:

int (1/x dx) = ln x + C, where C=an arbitrary constant

Then, if we substitute these, we still get the same basic result.

int (1/x dx)= 1 + int (1/x dx)

or

ln x +C1 = 1 + ln x +C2

Remember C1 is NOT necesarily equal to C2. We can then resolve your seemingly paradoxical sitaution by doing a little algebra.

C1 = C2 + 1, or
C2=C1-1

Substituting back:

ln x +C2 + 1= 1 + ln x +C2

Everything cancels out, so:

0=0

Answer 2

Sure...
ln x is an antiderivative of 1/x
ln x + 1 is also an antiderivative of 1/x
ln x - 4 is also an antiderivative of 1/x
ln x - 67 is also an antiderivative of 1/x
ln x + 111.2 is also an antiderivative of 1/x
ln x + e is also an antiderivative of 1/x
ln x - 6π is also an antiderivative of 1/x
ln x + 33φπ + φ² - 35√3is also an antiderivative of 1/x
ln x - 44.3423546762748574305642360743063416 is also an antiderivative of 1/x

In fact there is a family of functions ln x + C (C = constant) which is an antiderivative of 1/x, each of which DIFFER by a constant. This means that they are NOT equal.

^_^

Answer 3

differential coefficient of ln x=1/x and that of( ln x+1987) is also 1/x would you conclude 1987=0?same logic as integration is the inverse process of differentiation

Answer 4

remember the first rule of integration...plus C...it _is_ a matter of the constant.

Answer 5

Valuable discussion, just what I was searching for.

Answer 6

whatchutalkinbout, wills?

sorry, i couldn't resist the urge.