integral of 1/x

**acc100jt** · May 13th 2009, 07:57 PM

Hi,

I understand that differentiate both ln x and ln |x| will give 1/x.
But why $\int \frac{1}{x} dx=ln|x|+c$ .
Is it ok if I write $\int \frac{1}{x} dx=ln x+c$

Can someone help?

**Chris L T521** · May 13th 2009, 08:12 PM

Originally Posted by acc100jt

Hi,

I understand that differentiate both ln x and ln |x| will give 1/x.
But why $\int \frac{1}{x} dx=ln|x|+c$ .
Is it ok if I write $\int \frac{1}{x} dx=ln x+c$

Can someone help?

It depends on the interval x is defined on.

If $x\in\left(0,\infty\right)$ , then its safe to say that $\int\frac{1}{x}\,dx=\ln x+C$ . However, if $x\in\left(-\infty,0\right)$ , then $\int\frac{1}{x}\,dx=\ln\left(-x\right)+C$ .

Thus, it would be best to define $\int\frac{1}{x}\,dx=\ln\!\left|x\right|+C$ for $x\in\mathbb{R}\backslash\{0\}$ (any real number except zero)

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