Find the Maclaurin series for ln(1`+-` x) with these results and obtain the series representing ln((1+x)/(1-x)).

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Maclaurin series is defined as Taylor series in neighbourhood of 0 that is for function `f` Maclaurin series is

`f(x)=sum_(n=0)^infty (f^((n))(0))/(n!)x^n=f(0)+(f'(0))/1x+(f''(0))/2x^2+(f'''(0))/6x^3+(f^((4))(0))/24x^4+cdots`

Let's now calculate first few derivations of `ln(1+x)`.





So we have


Now we do the same thing for `ln(1-x)`





So we have


We can now calculate  Maclaurin series of `ln((1+x)/(1-x))`.


Hence we have

`ln((1+x)/(1-x))= x+x-(x^2)/2+(x^2)/2+(x^3)/3+(x^3)/3-(x^4)/4+(x^4)/4+cdots`

`ln((1+x)/(1-x))=2x+(2x^3)/2+(2x^5)/5+(2x^7)/7+cdots=sum_(n=1)^infty (2x^n)/n`  <--Solution

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