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

Ikunupe6v 2021-12-31 Answered
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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Expert Answer

Marcus Herman
Answered 2022-01-01 Author has 1875 answers
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
\(\displaystyle{x}={e}^{{u}},{u}={\ln{{x}}}\)
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Ronnie Schechter
Answered 2022-01-02 Author has 34 answers
Let's use your result :
\(\displaystyle\int{x}^{{a}}={\frac{{{x}^{{{a}+{1}}}}}{{{a}+{1}}}}+{C}={\frac{{{x}^{{{a}+{1}}}-{1}}}{{{a}+{1}}}}+{C}'\)
and take the limit as \(\displaystyle{a}\rightarrow-{1}\)
(since \(\displaystyle{x}^{{{a}+{1}}}={e}^{{{\left({a}+{1}\right)}{\ln{{\left({x}\right)}}}}}\)
\(\displaystyle\lim_{{{a}\rightarrow-{1}}}\int{x}^{{a}}{\left.{d}{x}\right.}={C}'+\lim_{{{a}\rightarrow-{1}}}{\frac{{{e}^{{{a}+{1}{\ln{{\left({x}\right)}}}}}-{1}}}{{{a}+{1}}}}={C}'+{\ln{{\left({x}\right)}}}\)
Hoping this will help your intuition,
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Vasquez
Answered 2022-01-09 Author has 8850 answers

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.

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