# 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)`.

`(ln(1+x))'=1/(1+x)`

`(1/(1+x))'=-1/(1+x)^2`

`(-1/((1+x)^2))'=2/(1+x)^3`

`(2/((1+x)^3))'=-6/(1+x)^4`

So we have

`ln(1+x)=x-(x^2)/2+(x^3)/3-(x^4)/4+(x^5)/5-cdots`

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

`(ln(1-x))'=-1/(1-x)`

`(-1/(1-x))'=-1/(1-x)^2`

`(-1/((1-x)^2))'=-2/(1-x)^3`

`(-2/((1-x)^3))'=-6/(1-x)^4`

So we have

`ln(1-x)=-x-(x^2)/2-(x^3)/3-(x^4)/4-(x^5)/5-cdots`

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

`ln((1+x)/(1-x))=ln(1+x)-ln(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**