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${\displaystyle f(x)=\ln(5-x)\,}$
${\displaystyle {\frac {d}{dx}}f(x)=-{\frac {1}{5-x}}=-{\frac {1}{1-(x-4)}}=-\sum _{n=0}^{\infty }(x-4)^{n}\,}$
${\displaystyle \int df(x)=\int \left(-\sum _{n=0}^{\infty }(x-4)^{n}\right)dx\,}$
${\displaystyle f(x)=C-\sum _{n=0}^{\infty }{\frac {(x-4)^{n+1}}{n+1}}\,}$
${\displaystyle f(x)=C-\sum _{n=1}^{\infty }{\frac {(x-4)^{n}}{n}}\,}$
${\displaystyle \ln(5-x)=C-\sum _{n=1}^{\infty }{\frac {(x-4)^{n}}{n}}\,}$
${\displaystyle \ln(5-4)=C-\sum _{n=1}^{\infty }{\frac {(4-4)^{n}}{n}}\,}$
${\displaystyle \ln(1)=C-\sum _{n=1}^{\infty }{\frac {(0)^{n}}{n}}\,}$
${\displaystyle \ln(1)=C\,}$
${\displaystyle 0=C\,}$
${\displaystyle f(x)=-\sum _{n=1}^{\infty }{\frac {(x-4)^{n}}{n}}\,}$