Finding the sum of the series $\sum_{n=0}^n \frac{x^{4n}}{4n+3}$ $$\sum_{n=0}^n \frac{x^{4n}}{4n+3}$$
Now we're given a hint to turn this into $x^{-3}\int g(t)dt$
I've gone ahead and done the following:
$$x^{-3}\sum_{n=0}^n \frac{x^{4n+3}}{4n+3}$$
And this turns into:
$$x^{-3}\sum_{n=0}^n \left(\int_0^x{t^{4n+2}}\right)$$
Now we know for a power series:
$$\sum_{n=0}^n q^{n}= \frac{1}{1-q}$$
And I need to turn the term from two lines ago into this form. 
$t^{4n+2}=t^{4n}{t^2}=\left((t^4)^{n}*t^2\right)$
Now is my thought process correct here?
$$x^{-3}\sum_{n=0}^n \left(\int_0^x{t^{4n+2}}\right)=x^{-3}t^2\sum_{n=0}^n \left(\int_0^x{t^{4n}}\right)=x^{-3}\int\left(\frac{t^2}{1-t^4}\right)$$
(And of course after this comes the fraction decomposition and the integral.)

Fine so far,
$$\sum_{n\geq 0}\frac{x^{4n+3}}{4n+3} = \int_{0}^{x}\frac{t^2}{1-t^4}\,dt = \frac{\text{arctanh}(x)-\arctan(x)}{2}$$
leads to:
$$ \sum_{n\geq 0}\frac{x^{4n}}{4n+3} = \frac{\text{arctanh}(x)-\arctan(x)}{2x^3}$$
for any $x\in(-1,1)$.
