f(x) = ln[(1+x^2)/(1-x^2)]
= ln(1+x^2) - ln(1-x^2)
i know log(1+x) = x^n+1 / n+1 but i dont know how i am supposed to modify the equation or whatever to get what i need...
$\displaystyle \ln (1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} + ... \mbox{ for }-1<t \leq 1$
Now to find,
$\displaystyle \ln (1+x^2)$ use $\displaystyle t=x^2 \mbox{ for }|x|<1$
And, $\displaystyle \ln (1-x^2) = \ln[(1-x)\cdot(1+x)] = \ln (1-x) + \ln (1+x)$
To find,
$\displaystyle \ln (1-x)$ use $\displaystyle t=-x \mbox{ for }|x|<1$
And combine them together.