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

\(\displaystyle{x}={e}^{{u}},{u}={\ln{{x}}}\)

\(\displaystyle{x}={e}^{{u}}\)

\(\displaystyle{\left.{d}{x}\right.}={e}^{{u}}{d}{u}\)

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

\(\displaystyle{x}={e}^{{u}},{u}={\ln{{x}}}\)