# Demystify integration of \int\frac{1}{x}dx I've le

Demystify integration of $$\displaystyle\int{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}$$
I've learned in my analysis class, that
$$\displaystyle\int{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}={\ln{{\left({x}\right)}}}$$
I can live with that, and it's what I use when solving equations like that. But how can I solve this, without knowing that beforehand.
Assuming the standard rule for integration is
$$\displaystyle\int{x}^{{a}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{a}+{1}}}}\cdot{x}^{{{a}+{1}}}+{C}$$
If I use that and apply this to $$\displaystyle\int{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}$$
$$\displaystyle\int{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}=\int{x}^{{-{1}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{1}}}{{-{1}+{1}}}}\cdot{x}^{{-{1}+{1}}}$$
$$\displaystyle={\frac{{{x}^{{0}}}}{{{0}}}}$$
Obviously, this doesn't work, as I get a division by 0. I don't really see, how I can end up with ln(x). There seems to be something very fundamental that I'm missing.
I study computer sciences, so, we usually omit things like in-depth math theory like that. We just learned that $$\displaystyle\int{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}={\ln{{\left({x}\right)}}}$$ and that's what we use.

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Marcus Herman
If you want to try to prove $$\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{x}}}}={\ln{{x}}}+{C}{\left(\ \text{ for >0}\right)}$$, try the substitution
$$\displaystyle{x}={e}^{{u}}$$
$$\displaystyle{\left.{d}{x}\right.}={e}^{{u}}{d}{u}$$
This substitution is justified because the exponential function is bijective from $$\displaystyle{\mathbb{{R}}}$$ to $$\displaystyle{\left({0},\infty\right)}$$ (hence for every x there exists a u) and continuously differentiable (which allows an integration by substitution).
$$\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{x}}}}=\int{\frac{{{e}^{{u}}{d}{u}}}{{{e}^{{u}}}}}={u}+{C}$$
Now just use the fact that natural log is the inverse of the exponential function. If
$$\displaystyle{x}={e}^{{u}},{u}={\ln{{x}}}$$
###### Not exactly what youâ€™re looking for?
Ronnie Schechter
$$\displaystyle\int{x}^{{a}}={\frac{{{x}^{{{a}+{1}}}}}{{{a}+{1}}}}+{C}={\frac{{{x}^{{{a}+{1}}}-{1}}}{{{a}+{1}}}}+{C}'$$
and take the limit as $$\displaystyle{a}\rightarrow-{1}$$
(since $$\displaystyle{x}^{{{a}+{1}}}={e}^{{{\left({a}+{1}\right)}{\ln{{\left({x}\right)}}}}}$$
$$\displaystyle\lim_{{{a}\rightarrow-{1}}}\int{x}^{{a}}{\left.{d}{x}\right.}={C}'+\lim_{{{a}\rightarrow-{1}}}{\frac{{{e}^{{{a}+{1}{\ln{{\left({x}\right)}}}}}-{1}}}{{{a}+{1}}}}={C}'+{\ln{{\left({x}\right)}}}$$
Vasquez

To show $$\int\frac{dx}{x}=\ln x+C$$ for positive x, we can show that the derivative of $$\ln x$$ is $$\frac{1}{x}$$. This can be done using the standard limit
$$\lim_{x\rightarrow0}\frac{\ln(1+x)}{x}=1$$
and the definition of the derivative. We have
$$\frac{d}{dx}(\ln x)=\lim_{h\rightarrow0}\frac{\ln(x+h)-\ln x}{h}=\lim_{h\rightarrow0}\frac{1}{h}\ln(1+\frac{h}{x})=[t=\frac{1}{x}]=\lim_{t\rightarrow0}\frac{1}{x}$$
$$\cdot\frac{\ln(1+t)}{t}=\frac{1}{x}$$
which is what we wanted to show.