1. ## power series again

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...

2. Originally Posted by jeph
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...
$\ln (1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} + ... \mbox{ for }-1

Now to find,
$\ln (1+x^2)$ use $t=x^2 \mbox{ for }|x|<1$

And, $\ln (1-x^2) = \ln[(1-x)\cdot(1+x)] = \ln (1-x) + \ln (1+x)$

To find,
$\ln (1-x)$ use $t=-x \mbox{ for }|x|<1$

And combine them together.