Integration math question

You're missing an integration constant.

 

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

Please tell me if I am wrong.

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