Is in the integral of 1/x ln(x)+c or ln(x)+ln(c)?

Question

I have seen it written the first way in some problems and the second in others. All have had correct answers. Can someone please explain why this is? Thank!

Answer 1

The c is just an unknown constant.  It can be expressed in terms of another unknown constant, as long as the range of values is the same.  The constant c is somewhere between -∞ and +∞, but so is ln(b), so it is acceptable to write the constant of integration c in terms of the natural logarithm of another constant b.

 

c = ln(b)

 

An example helps illustrate why it is convenient to do this.  Suppose you have

 

dy/dx = ky

 

Then

 

dy/y = kdx

 

∫dy/y = ∫kdx

 

ln(y) = kx + c

 

e^ln(y) = e^ce^kx

 

y = e^ce^kx

 

But e^c is still just a constant.  You can call it A.

 

y = Ae^kx

 

But if we express the integration constant as ln(c), we get

 

ln(y) = kx + ln(c)

 

y = ce^kx

 

Again, there is a constant to be determined in front of e^kx.  In one case we called it A, in another we called it c.  Either way, it turns out to be y(0), so it gives the same final result.

 

y(x) = y(0) e^kx

 

Does this help you see why using either c or ln(b) as the integration constant works?  It looks like there are fewer steps if you use ln(b).