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For a function^{$f\left(x\right)$} the Taylor series expansion about a point ^x₀ is given by,$f(x-x_0)=f\left(x_0\right)+f\text{'}\left(x_0\right).(x-x_0)+f\text{'}\text{'}\left(x_0\right).\frac{{(x-x_0)}^2}{2!}+f\text{'}\text{'}\text{'}\left(x_0\right).\frac{{(x-x_0)}^3}{3!}+...$

Rewrite $\frac{1}{x}$by adding and subtracting 1 from the denominator, as follows:

$\frac{1}{x}=\frac{1}{1+(x-1)}$and$\frac{1}{1-s}=\underset{n=0}{\overset{鈭�}{鈭�}}s\text{'}\text{'}$

It follows,

$$\begin{gathered} \frac{1}{x}=\frac{1}{1+(x-1)} \\ =\underset{n=1}{\overset{鈭�}{鈭�}}{(1-x)}^n \\ =\underset{n=1}{\overset{鈭�}{鈭�}}{{(-1)}^n(1-x)}^n \end{gathered}$$

Therefore, the Taylor series for$\frac{1}{x}$ around x₀=1 is given by: .$\frac{1}{x}$role="math" localid="1664187564177" $=\underset{n=1}{\overset{鈭�}{鈭�}}{(-1)}^n(1-x)$

Substitute $\frac{1}{x}=\underset{n=1}{\overset{鈭�}{鈭�}}{(-1)}^n{(1-x)}^n$in the equation role="math" localid="1664187704795" $\ln x={鈭�}_1^x\frac{1}{t}dt,$we get,

$\ln x={鈭�}_1^x\underset{n=1}{\overset{鈭�}{鈭�}}{(-1)}^n{(t-1)}^{n}dt$

Interchange the integral and the sum,

role="math" localid="1664188068704" $\ln x=\underset{n=1}{\overset{鈭�}{鈭�}}{鈭�}_1^x{(-1)}^n{(t-1)}^{n}dt$

$\ln x=\underset{n=1}{\overset{鈭�}{鈭�}}{(-1)}^n{鈭�}_1^x{(t-1)}^{n}dt$

Take integration,

$\ln x=\underset{n=1}{\overset{鈭�}{鈭�}}{(-1)}^n{{\overline{)\frac{{(t-1)}^{n+1}}{n+1}}}^x}_1$

Compute the boundaries,

$\ln x=\underset{n=1}{\overset{鈭�}{鈭�}}{(-1)}^n\frac{{(t-1)}^{n+1}}{n+1}$

Let,

k = n + 1

n = k - 1

So,

$\ln x=\underset{\mathrm{k}=1}{\overset{鈭�}{鈭�}}{(-1)}^{\mathrm{k}-1}\frac{{(\mathrm{x}-1)}^k}{k}$

Now substitute k for n in the above equation, it is only a dummy index the name is not important, we need to respect the range from 1 to$鈭�$.