<s>[INST] If $f(x)=\lim_{n\to\infty}[2x+4x^3+\cdots+2nx^{2n-1}]$, $0
If $f(x)=\lim_{n\to\infty}[2x+4x^3+\cdots+2nx^{2n-1}]$, $0<x<1$, then find $\int f(x)\mathrm{d}x$
$$f(x)=\lim_{n\to\infty}2x[1+2x^2+\cdots+nx^{2n}]$$
$$S=\frac{f(x)}{2x}=1+2x^2+\cdots+nx^{2n}$$
$S$ is an AGP. I used the general method for finding $S$.
$$x^2S=x^2+2x^4+\cdots+nx^{2n+2}$$
$$(1-x^2)S=1+(x^2+x^4+\cdots+x^{2n})+nx^{2n+2}$$
$$(1-x^2)S=\frac{1-x^{2n+1}}{1-x^2}+nx^{2n+2}$$
$$S=\frac{1-x^{2n+1}}{(1-x^2)^2}+\frac{nx^{2n+2}}{1-x^2}$$
$$f(x)=\lim_{n\to\infty}\left(2x\frac{1-x^{2n+1}}{(1-x^2)^2}+2x\frac{nx^{2n+2}}{1-x^2}\right)$$
I got stuck here.
 [/INST] You can turn that long expression into a infinite series.
$$ f(x) = \sum_{n=1}^\infty 2n x^{2n-1}
$$
Let $g = \int f(x)\,dx$. By termwise integration, (This is justified by differentiating $g$ and noticing that $g$ has the same radius of convergence as $f$. Then it follows from the uniqueness of antiderivative up to a constant.)
$$ g(x) = \sum_{n=1}^\infty 2n \frac{x^{2n}}{2n} = \sum_{n=1}^\infty x^{2n} = \sum_{n=1}^\infty (x^2)^n = \frac{x^2}{1-x^2}
$$
and finally add the constant of integration to the end.
</s>