Find the Taylor Series expansion of $\tanh^{-1} x = \int_0^x \frac{1}{1-t^2} dt$
First, I found the taylor series expansion for $\frac{1}{1-x} = 1 +x+x^2 + x^3 + ...$
I substitute $x=t^2$ above,
$\frac{1}{1-t^2} = 1 + t^2 + t^4 + t^6 + ... = \sum_{n=0} t^{2n}$
But this expansion is wrong,
The correct answer is: $\tanh^{-1} = \sum_{n=0} \frac{x^{2n+1}}{2n+1}$
$$\frac {1}{1-t^2}＝\sum_{k＝0}^{}t^{2k}\implies\int_{0}^{x}\frac {1}{1-t^2}dt＝\int_{0}^{x}(\sum_{k＝0}^{}t^{2k})dt＝\sum_{k＝0}^{}\int_{0}^{x}t^{2k}dt＝\sum_{k＝0}^{}\frac {x^{2k＋1}}{2k＋1}$$
And therefore : $$\tanh^{-1}(x)＝\sum_{k＝0}^{}\frac {x^{2k＋1}}{2k＋1}$$