Discrepancy in Integration of 1/x by Parts

[Gear300]

If we were to integrate 1/x by parts using u = 1/x and dv = 1dx, then it would end up as:
Int(dx/x) = 1 + Int(dx/x), ending up with 1 = 0. Why was there a discrepancy in the integration?
--Int() refers to integral

 

[morphism]

You're missing an integration constant.

 

[exk]

\int\frac{1}{x}dx=uv -\int vdu= \frac{x+c}{x} - \int \frac{-(x+c)}{x^{2}}dx = etc...
u=\frac{1}{x}
du=\frac{-1}{x^{2}}
dv=1dx
v=x+c

Please tell me if I am wrong.

Regardless, you are looking for the result to be ln|x|+c.

 

[Defennder]

v=x+c

I don't think there should be a '+ c' in this intermediate step. Anyway, integrating
1/x gives ln(x) + c by definition. Some books actually define ln(x) to be the anti-derivative of that.